A New Method of Investigating the Sums of Infinite Series. By the Rev. Samuel Vince, A. M. F. R. S.

XX. A new Method of investigating the Sums of Infinite Series. By the Rev. Samuel Vince, A. M. F. R. S. Read June 2, 1791. The summation of infinite series is a subject, not only of curious speculation, but also of the greatest importance in the various branches of mathematics and philosophy; in consequence of which it has always claimed a very considerable share of attention from the most celebrated mathematicians. I shall therefore make no apology for offering to the public the following new and very expeditious method, by which we may obtain the sums of a great variety of series, most of which have never before been treated of. As the summation depends on the sums of the reciprocals of the powers of the natural numbers, tables of such sums are given as far as the 40th power to twelve places of decimals, by which the sums of the series will be found true to ten or eleven places; and if greater accuracy were required (which is a case that can very rarely happen) it might easily be obtained by continuing the tables. The first and third columns shew the sums, and the second and fourth the powers corresponding. TABLE Mr. Vince on the Sums TABLE I. Sum of $\frac{1}{2^n} + \frac{1}{3^n} + \frac{1}{4^n} + \ldots$ ad infinitum. | Sum | n | |-----|---| | A = ,644934066848 | 2 | | B = ,202056903159 | 3 | | C = ,082323233711 | 4 | | D = ,036927755107 | 5 | | E = ,017343061984 | 6 | | F = ,008349277387 | 7 | | G = ,004077356198 | 8 | | H = ,002008392826 | 9 | | I = ,000994575128 | 10 | | K = ,000494188604 | 11 | | L = ,000246086553 | 12 | | M = ,000122713347 | 13 | | N = ,000061248135 | 14 | | O = ,000030588236 | 15 | | P = ,000015282259 | 16 | | Q = ,000007637196 | 17 | | R = ,000003817292 | 18 | | S = ,000001908212 | 19 | | T = ,000000953961 | 20 | | V = ,000000476932 | 21 | | Sum | n | |-----|---| | W = ,000000238450 | 22 | | X = ,00000019219 | 23 | | Y = ,000000059608 | 24 | | Z = ,000000029803 | 25 | | A' = ,000000014901 | 26 | | B' = ,000000007450 | 27 | | C' = ,000000003725 | 28 | | D' = ,000000001863 | 29 | | E' = ,000000000931 | 30 | | F' = ,000000000465 | 31 | | G' = ,000000000233 | 32 | | H' = ,000000000116 | 33 | | I' = ,000000000058 | 34 | | K' = ,000000000029 | 35 | | L' = ,000000000015 | 36 | | M' = ,000000000007 | 37 | | N' = ,000000000004 | 38 | | O' = ,000000000002 | 39 | | P' = ,000000000001 | 40 | TABLE II. Sum of $\frac{1}{2^n} - \frac{1}{3^n} + \frac{1}{4^n} - \frac{1}{5^n} + \&c.$ ad infinitum. | Sum | $n$ | |-----|-----| | $a = ,177532966576$ | 2 | | $b = ,098457322630$ | 3 | | $c = ,052967170503$ | 4 | | $d = ,027880229587$ | 5 | | $e = ,014448908703$ | 6 | | $f = ,007406180072$ | 7 | | $g = ,003766998147$ | 8 | | $h = ,001905702459$ | 9 | | $i = ,000960492403$ | 10 | | $k = ,000482856502$ | 11 | | $l = ,000242314856$ | 12 | | $m = ,000121457237$ | 13 | | $n = ,000060829654$ | 14 | | $o = ,000030448787$ | 15 | | $p = ,000015235790$ | 16 | | $q = ,000007621788$ | 17 | | $r = ,000003812130$ | 18 | | $s = ,000001906491$ | 19 | | $t = ,000000953389$ | 20 | | $v = ,0000000476742$ | 21 | | Sum | $n$ | |-----|-----| | $w = ,000000238386$ | 22 | | $x = ,000000119199$ | 23 | | $y = ,000000059602$ | 24 | | $z = ,000000029801$ | 25 | | $a' = ,000000014901$ | 26 | | $b' = ,000000007450$ | 27 | | $c' = ,000000003725$ | 28 | | $d' = ,000000001863$ | 29 | | $e' = ,000000000931$ | 30 | | $f' = ,000000000465$ | 31 | | $g' = ,000000000233$ | 32 | | $h' = ,000000000116$ | 33 | | $i' = ,000000000058$ | 34 | | $k' = ,000000000029$ | 35 | | $l' = ,000000000015$ | 36 | | $m' = ,000000000007$ | 37 | | $n' = ,000000000004$ | 38 | | $o' = ,000000000002$ | 39 | | $p' = ,000000000001$ | 40 | Mr. Vince on the Sums TABLE III. Sum of $\frac{1}{2^n} + \frac{1}{4^n} + \frac{1}{6^n} + \&c.$ ad infinitum. | Sum | n | |-----|---| | A'' = ,411233516712 | 2 | | B'' = ,100257112895 | 3 | | C'' = ,067645202107 | 4 | | D'' = ,032403992347 | 5 | | E'' = ,015895985344 | 6 | | F'' = ,007877728730 | 7 | | G'' = ,003922177173 | 8 | | H'' = ,001957047643 | 9 | | I'' = ,000977533765 | 10 | | K'' = ,000488522553 | 11 | | L'' = ,000244200705 | 12 | | M'' = ,000122085292 | 13 | | N'' = ,000061038895 | 14 | | O'' = ,000030518512 | 15 | | P'' = ,000015259024 | 16 | | Q'' = ,000007629452 | 17 | | R'' = ,000003814712 | 18 | | S'' = ,000001907352 | 19 | | T'' = ,000000953675 | 20 | | V'' = ,000000476837 | 21 | | Sum | n | |-----|---| | W'' = ,000000238419 | 22 | | X'' = ,000000119209 | 23 | | Y'' = ,000000059605 | 24 | | Z'' = ,000000029502 | 25 | | A''' = ,000000014901 | 26 | | B''' = ,000000007450 | 27 | | C''' = ,000000003725 | 28 | | D''' = ,000000001863 | 29 | | E''' = ,000000000931 | 30 | | F''' = ,000000000465 | 31 | | G''' = ,000000000233 | 32 | | H''' = ,000000000116 | 33 | | I''' = ,000000000058 | 34 | | K''' = ,000000000029 | 35 | | L''' = ,000000000015 | 36 | | M''' = ,000000000007 | 37 | | N''' = ,000000000004 | 38 | | O''' = ,000000000002 | 39 | | P''' = ,000000000001 | 40 | ### TABLE IV. Sum of $\frac{1}{3^n} + \frac{1}{5^n} + \frac{1}{7^n} + \&c.$ ad infinitum. | Sum | n | |-----|---| | $a'' = ,233700550136$ | 2 | | $b'' = ,051799790264$ | 3 | | $c'' = ,014678031604$ | 4 | | $d'' = ,004523762760$ | 5 | | $e'' = ,001447076640$ | 6 | | $f'' = ,000471548657$ | 7 | | $g'' = ,000155179025$ | 8 | | $h'' = ,000051345183$ | 9 | | $i'' = ,000017041362$ | 10 | | $k'' = ,000005666051$ | 11 | | $l'' = ,000001885848$ | 12 | | $m'' = ,000000628055$ | 13 | | Sum | n | |-----|---| | $n'' = ,000000209240$ | 14 | | $o'' = ,00000069724$ | 15 | | $p'' = ,00000023234$ | 16 | | $q'' = ,00000007744$ | 17 | | $r'' = ,00000002531$ | 18 | | $s'' = ,00000000864$ | 19 | | $t'' = ,00000000286$ | 20 | | $v'' = ,00000000095$ | 21 | | $w'' = ,00000000032$ | 22 | | $x'' = ,00000000011$ | 23 | | $y'' = ,00000000004$ | 24 | | $z'' = ,00000000001$ | 25 | --- **PROP. I.** To find the sum of the sums of the reciprocal squares, cubes, &c. &c. ad infinitum. By division $\frac{1}{x-1 \times x} = \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4} + \&c.$ ad inf.; hence if we make each of these terms the general term of a series, and write 2, 3, 4, &c. ad inf. for $x$, we have $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \&c. = (\text{Table I.}) A + B + C + D + \&c.$; but $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \&c. \text{ad inf.} = 1$; hence $A + B + C + D + \&c. \text{ad inf.} = 1$. As \(\frac{1}{x \times x + 1} = \frac{1}{x^2} - \frac{1}{x^3} + \frac{1}{x^4} - \frac{1}{x^5} + \&c.\ ad inf.\); we have, by the same method of proceeding, \(A - B + C - D + \&c.\ ad inf. = \frac{1}{2}\); consequently \(A + C + E + \&c. = \frac{3}{4}\), and \(B + D + F + \&c. = \frac{1}{4}\). Because \(\frac{1}{x - 1 \times x} = \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4} + \&c.\ ad inf.\); if for \(x\) we write 2, 4, 6, \&c., then will \(\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \&c. = (\text{Tab. 3}) A'' + B'' + C'' + D'' + \&c.;\) but \(\frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \&c. = \text{hyp. log. 2};\) hence \(A'' + B'' + C'' + D'' + \&c. = \text{hyp. log. 2}.\) If in the same expression we write 3, 5, 7, \&c. for \(x\), then \(\frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + \&c. = (\text{Tab. 4.}) a'' + b'' + c'' + \&c.;\) but \(\frac{1}{2 \cdot 3} + \frac{1}{4 \cdot 5} + \frac{1}{6 \cdot 7} + \&c. = 1 - \text{hyp. log. 2};\) hence \(a'' + b'' + c'' + \&c. = 1 - \text{hyp. log. 2}.—Hence from either of these two last cases, we have a very expeditious method of finding the hyp. log. 2. **Prop. II.** To find the sum of the infinite series whose general term is \(\frac{1}{mx^r \pm n}.\) By division \(\frac{1}{mx^r \pm n} = \frac{1}{mx^r} = \frac{n}{m^2x^{2r}} + \frac{n^2}{m^3x^{3r}} + \frac{n^3}{m^4x^{4r}} + \&c.\ ad inf.;\) hence, if \(\frac{1}{mx^r \pm n}\) be made the general term of a series, and for \(x\) we write 2, 3, 4, \&c., its sum will be equal to the sums of another set of serieses, whose terms are the powers of the reciprocals of the natural numbers respectively multiplied into into $\frac{1}{m}$, $\frac{n}{m^2}$, $\frac{n^2}{m^3}$, &c.; hence the sum of each of these series being known from the tables, the sum of the given series will be found. Ex. 1. Let $\frac{1}{x^2 + 1}$ be the general term; now $\frac{1}{x^2 + 1} = \frac{1}{x^2} - \frac{1}{x^4} + \frac{1}{x^6} - \frac{1}{x^8} + \&c.$; hence if for $x$ we write 2, 3, 4, &c. we have $\frac{1}{5} + \frac{1}{10} + \frac{1}{17} + \frac{1}{26} + \&c. = A - C + E - G + \&c. = (\text{by Tab. 1.})$ ,576674037469. Ex. 2. Let $\frac{1}{x^2 - 1}$ be the general term; then, by the same method of proceeding, $\frac{1}{3} + \frac{1}{8} + \frac{1}{15} + \frac{1}{24} + \&c. = A + C + E + \&c. = (\text{by Prop. 1.}) \frac{3}{4}$. Cor. Because $\frac{1}{8} + \frac{1}{24} + \frac{1}{48} + \&c. = \frac{1}{8} \times 1 + \frac{1}{3} + \frac{1}{6} + \&c. = (\text{as } 1 + \frac{1}{3} + \frac{1}{6} + \&c. \text{ is the reciprocal of the figurative numbers of the second order}) \frac{1}{8} \times 2 = \frac{1}{4}$; therefore $\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \&c. = \frac{1}{2}$. Also, as $\frac{1}{x^2 - 1} = \frac{1}{x^2} + \frac{1}{x^4} + \frac{1}{x^6} + \&c.$; if we write 2, 4, 6, &c. for $x$, we have $\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \&c. = (\text{by Tab. 3.}) A'' + C'' + E'' + \&c. = \frac{1}{2}$; but, by Prop. 1. $A'' + B'' + C'' + D'' + \&c. = \text{hyp. log. 2}$; hence $B'' + D'' + F'' + \&c. = -\frac{1}{2} + \text{hyp. log. 2}$. Ex. 3. Let the general term be $\frac{1}{x^3 - 1} = \frac{1}{x^3} + \frac{1}{x^6} + \frac{1}{x^9} + \&c.$, and, by writing 2, 3, 4, &c. for $x$, we have $\frac{1}{7} + \frac{1}{26} + \frac{1}{63} + \&c. = B + E + H + \&c. = ,221689395104$. Ex. 4. Let the general term be \( \frac{1}{3x^2 - 2} = \frac{1}{3x^2} + \frac{2}{9x^3} + \frac{4}{27x^4} + \&c. \), and, by writing 2, 3, 4, &c. for \( x \), &c. we have \( \frac{1}{46} + \frac{1}{241} + \frac{1}{766} + \&c. \). \[ \frac{1}{3}C + \frac{2}{9}G + \frac{4}{27}L + \&c. = ,028385252052. \] Ex. 5. To find the sum of the series \( \frac{1}{9} - \frac{1}{26} + \frac{1}{65} - \frac{1}{124} + \&c. \). If we write 2, -3, 4, -5, &c. for \( x \), the general term will be \( \frac{1}{x^3 + 1} = \frac{1}{x^3} - \frac{1}{x^6} + \frac{1}{x^9} - \frac{1}{x^{12}} + \&c. \). Now, by writing 2, -3, 4, -5, &c. for \( x \), the serieses of which \( \frac{1}{x^3}, \frac{1}{x^6}, \&c. \) are the general terms, will be alternately + and -, and therefore their sums will be found in Tab. 2. and the serieses of which \( \frac{1}{x^9}, \frac{1}{x^{12}}, \&c. \) are the general terms will have their terms all +, and therefore their sums will be found in Tab. 1. Hence the sum required \( = b + b + o + \&c. - E - L - R - \&c. = ,082800931803. \) PROP. III. To find the sum of the sums of the reciprocals of the odd powers in Tab. 2. By division \( \frac{1}{x-1 \times x} = \frac{1}{x^2} + \frac{1}{x^3} + \frac{1}{x^4} + \frac{1}{x^5} + \&c. \); hence by writing 2, -3, 4, -5, &c. for \( x \), the sums of the serieses of which \( \frac{1}{x^3}, \frac{1}{x^6}, \&c. \) are the general terms, may be found by Tab. 2. and the other sums by Tab. 1.; hence \( \frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \&c. = A + C + E + \&c. + b + d + f + \&c.; \) but \( \frac{1}{1 \cdot 2} + \frac{1}{3 \cdot 4} + \frac{1}{5 \cdot 6} + \&c. \) &c. = -\frac{1}{2} + 2 \text{ hyp. log. } 2; \text{ and by Prop. I. } A + C + E + &c. = \frac{3}{4}; \text{ hence } b + d + f + &c. = -\frac{5}{4} + 2 \text{ hyp. log. } 2. **PROP. IV.** To find the sum of the infinite series whose general term is \[ \frac{x^s}{mx^r \pm n}. \] By division \[ \frac{x^s}{mx^r \pm n} = \frac{1}{mx^r} = \frac{n}{m^2x^{2r}} + \frac{n^2}{m^3x^{3r}} + &c. \text{ ad inf.}; \] hence the sum of the series of which \[ \frac{x^s}{mx^r \pm n} \] is the general term, is found as in Prop. 2. Here \( r \) must be greater than \( s \) at least by 2, otherwise the sum will be infinite. **Ex. 1.** Let the general term be \[ \frac{x^2}{x^4 + 1} = \frac{1}{x^2} - \frac{1}{x^5} + \frac{1}{x^8} - &c.; \] hence if for \( x \) we write 2, 3, 4, &c. we have \[ \frac{4}{17} + \frac{9}{82} + \frac{16}{257} + &c. = A - E + I - N + &c. = 538527924723.\] If for \( x \) we write 2, 4, 6, &c. we get \[ \frac{4}{17} + \frac{16}{257} + \frac{36}{1296} + &c. = A'' - E'' + I'' - N'' + &c. = 396257616555. \] **Ex. 2.** Let the general term be \[ \frac{x}{3x^3 - 1} = \frac{1}{3x^2} + \frac{1}{9x^5} + \frac{1}{27x^8} + &c.; \] hence if we write 2, 3, 4, &c. for \( x \), we have \[ \frac{2}{23} + \frac{3}{80} + \frac{4}{191} + &c. = \frac{1}{3} A + \frac{1}{9} D + \frac{1}{27} G + &c. = 219238483448. \] By this proposition we may find the sum of any series whose general term is \[ \frac{ax^s + bx^{s-1} + cx^{s-2} + &c.}{mx^r \pm n}; \] for this resolves itself into \[ \frac{ax^s}{mx^r \pm n}, \frac{bx^{s-1}}{mx^r \pm n}, &c. &c., \text{ the sum of each of which series is found.} \] found by this proposition. Now the \( s+1 \)th differences of the numerators of this general term are = 0, and therefore it comprehends all series under such circumstances. For example, let the given series be \( \frac{4}{17} + \frac{13}{82} + \frac{26}{257} + \frac{43}{626} \). Here the third differences of the numerators = 0; to find therefore the general expression for the numerator, assume \( ax^2 + bx + c \) for it; and, by writing 2, 3, 4, for \( x \), we have \( 4a + 2b + c = 4, 9a + 3b + -c = 13, 16a + 4b + c = 26 \); hence \( a = 2, b = -1, c = -2 \); and as the denominator is manifestly \( x^4 + 1 \), the general term will be \( \frac{2x^2 - x - 2}{x^4 + 1} = \frac{2x^2}{x^4 + 1} - \frac{x}{x^4 + 1} - \frac{2}{x^4 + 1} \), each of which being made the general term of a series, their sum will be found to be respectively 1,077055849446, 0,194173022145, and 0,156955159332; hence the sum of the given series is 0,725927667969. If \( s \) be negative, the general term becomes \( \frac{1}{x^i \times mx^r \pm n} = \frac{1}{mx^i + s} \) \( = \frac{1}{m^i x^{2r + i}} + \frac{1}{m^i x^{3r + i}} + &c. \) Ex. 1. To find the sum of \( \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} - &c. \) ad inf. Here the general term is \( \frac{1}{x - 1 \times x \times x + 1} = \frac{1}{x \times x^2 - 1} = \frac{1}{x^3 + \frac{1}{x^5} + \frac{1}{x^7} + &c.} \); hence, by writing 2, -3, 4, -5, &c. for \( x \), we have the sum = \( b + a + f + &c. = (\text{by Prop. 3.}) - \frac{5}{4} + 2 \text{ hyp. log. } 2 \). If \( \frac{1}{x - 1 \times x^3 \times x + 1} \) be the general term it resolves itself into \( \frac{1}{x^5} + \frac{1}{x^7} + \frac{1}{x^9} + &c. \); consequently the sum of \( \frac{1}{1 \cdot 2^3 \cdot 3} - \frac{1}{2 \cdot 3^3 \cdot 4} + &c. \) \[ \frac{1}{3 \cdot 4^5 \cdot 5} + \text{&c.} = -b - \frac{5}{4} + 2 \text{ hyp. log. } 2. \] In like manner the sum of \[ \frac{1}{1 \cdot 2^5 \cdot 3} - \frac{1}{2 \cdot 3^5 \cdot 4} + \frac{1}{3 \cdot 4^5 \cdot 5} + \text{&c.} = -b - d - \frac{5}{4} + 2 \text{ hyp. log. } 2. \] Thus we may proceed as far as we please by adding two powers to the middle term; and hence this remarkable property of the serieses, that the difference of the sums of the serieses where the middle term is \(x, x^3, x^5, \text{&c.}\) is \(b, d, f, \text{&c.}\) respectively. Ex. 2. In like manner if the general term be \[ \frac{1}{x-1 \times x^3 \times x+1}, \] and we write 2, 3, 4, &c. for \(x\), we have \[ \frac{1}{1 \cdot 2^3 \cdot 3} + \frac{1}{2 \cdot 3^3 \cdot 3} + \frac{1}{3 \cdot 4^3 \cdot 5} + \text{&c.} = D+F+H+\text{&c.} = (\text{by Prop. I.}) \frac{1}{4} - B. \] Hence also \[ \frac{1}{1 \cdot 2^3 \cdot 3} + \frac{1}{2 \cdot 3^3 \cdot 4} + \text{&c.} = \frac{1}{4} - B - D; \text{ and so on as before.} \] If the general term be under the form \[ \frac{1}{x^n \cdot x+m}, \] it will be most convenient to resolve it thus: by division \[ \frac{1}{x+m} = \frac{1}{x} - \frac{m}{x^2} + \frac{m^2}{x^3} - \text{&c.} = \frac{m^n}{x^n \cdot x+m}; \text{ hence } \frac{1}{x^n \cdot x+m} = \] \[ \frac{1}{x+m} - \frac{1}{x} + \frac{m}{x^2} - \frac{m^2}{x^3} + \text{&c.} \times \frac{1}{m^n} = -\frac{m}{x \cdot x+m} + \frac{m^2}{x^2 \cdot x^3} + \text{&c.} \times \frac{1}{m^n}, \] where the sign on the left hand will be + or − according as \(n\) is even or odd, and the number of terms on the right is \(= n\). Now the sum of the series whose general term is \[ \frac{m}{x \cdot x+m} \] is well known, and the sums of the other are found from the tables. Ex. 1. To find the sum of \[ \frac{1}{2^2 \cdot 3} + \frac{1}{3^2 \cdot 4} + \frac{1}{4^2 \cdot 5} + \text{&c. ad inf.} \] Here the general term is \(\frac{1}{x^2 \times x + 1} = -\frac{1}{x \cdot x + 1} + \frac{1}{x^2}\), and by writing \(2, 3, 4, \&c.\) for \(x\), we have the sum \(= -\frac{1}{2 \cdot 3} - \frac{1}{3 \cdot 4} - \&c. + A = -\frac{1}{2} + A\). In like manner \(\frac{1}{2^3 \cdot 3} + \frac{1}{4^2 \cdot 5} + \frac{1}{6 \cdot 7} + \&c. = -1 + \text{hyp.log.} z + A''\). Also \(\frac{1}{2^3 \cdot 5} + \frac{1}{3^3 \cdot 6} + \frac{1}{4^3 \cdot 7} + \&c. = \frac{13}{12} - 3A + 9B \times \frac{1}{27}\). If \(m\) be negative, then \(\frac{1}{x^n \cdot x - m} = \frac{m}{x \cdot x - m} - \frac{m^2}{x^2} - \&c. \times \frac{1}{m^2}\). Hence \(\frac{1}{2^4 \cdot 1} + \frac{1}{3^4 \cdot 2} + \frac{1}{4^4 \cdot 3} + \&c. = 1 - A - B - C\); and so on for others of the same kind. If the general term be under this form \(\frac{1}{x^n \cdot ax^n + m}\), then, in like manner, we have \(\pm \frac{1}{x^n \cdot ax^n + m} = \frac{1}{ax^n + m} - \frac{1}{ax^n} + \frac{m}{a^2x^{2n}} - \&c. \times \frac{ar}{m^r}\), where the sign on the left hand will be \(+\) or \(-\), according as \(r\) is even or odd, and the number of terms on the right is \(= r + 1\). Ex. 1. To find the sum of \(\frac{1}{2^4 \cdot 5} + \frac{1}{3^4 \cdot 10} + \frac{1}{4^4 \cdot 17} + \&c.\) Here \(m = 1, n = 2, r = 2, a = 1\), and the general term \(\frac{1}{x^2 \times x^2 + 1} = \frac{1}{x^2 + 1} - \frac{1}{x^2} + \frac{1}{x^4}\); now the sum of the series whose general term is \(\frac{1}{x^2 + 1}\) is \(= ,576674037469\), by Prop. 2.; consequently the sum required \(= ,576674037469 - A + C = ,014063204332\). Ex. 2. If the given series be \(\frac{1}{4 \cdot 5} + \frac{1}{9 \cdot 10} + \frac{1}{16 \cdot 17} + \&c.\) the general general term will be \(\frac{1}{x^2 \cdot x + 1} = -\frac{1}{x^2 + 1} + \frac{1}{x^2}\); hence, by writing \(2, 3, 4, \&c.\) for \(x\), we have the sum \(= -576674037469 + A = 06826002938\). If \(m\) be negative, then \(\frac{1}{x^n \cdot ax^n - m} = \frac{1}{ax^n - m} - \frac{m}{a^2 x^{2n}} - \&c.\) Ex. 1. To find the sum of \(\frac{1}{1 \cdot 2^2 \cdot 3} + \frac{1}{2 \cdot 3^2 \cdot 4} + \frac{1}{3 \cdot 4^2 \cdot 5} + \&c.\) Here the general term is \(\frac{1}{x^2 \cdot x^2 \cdot x + 1} = \frac{1}{x^2 \cdot x^2 - 1} = \frac{1}{x^2 - 1} - \frac{1}{x^2}\); now, by writing \(2, 3, 4, \&c.\) for \(x\), the sum of the series whose general term is \(\frac{1}{x^2 - 1}\) is \(= \frac{3}{4}\), by Prop. 2.; hence the sum required \(= \frac{3}{4} - A\). Ex. 2. Let the given series be \(\frac{1}{1 \cdot 2^2 \cdot 3} + \frac{1}{3 \cdot 4^2 \cdot 5} + \frac{1}{5 \cdot 6^2 \cdot 7} + \&c.\). Here the general term is the same as before, writing \(2, 4, 6, \&c.\) for \(x\); and, by Prop. 2. the sum of the series whose general term is \(\frac{1}{x^2 - 1}\) is \(= \frac{1}{2}\); hence the sum \(= \frac{1}{2} - A''\). Ex. 3. In like manner the sum of the series \(\frac{1}{1 \cdot 2^2 \cdot 3} + \frac{1}{2 \cdot 3^2 \cdot 4} + \frac{1}{3 \cdot 4^2 \cdot 5} + \&c. = 221689395104 - B\). Ex. 4. To find the sum of \(\frac{1}{3 \cdot 4^2 \cdot 5} + \frac{1}{8 \cdot 9^2 \cdot 10} + \frac{1}{15 \cdot 16^2 \cdot 17} + \&c.\). Here the general term is \(\frac{1}{x^2 - 1 \cdot x^4 \cdot x^2 + 1} = \frac{1}{x^4 \cdot x^4 - 1} = \frac{1}{x^4 - 1} - \frac{1}{x^4}\); but the sum of the series whose general term is \(\frac{1}{x^4 - 1}\) is \(= 086662976264\); hence the sum required \(= 086662976264 - C\). S f 2 PROP. PROP. V. To find the sum of the infinite series \( \frac{1}{15} + \frac{1}{40} + \frac{1}{85} + \frac{1}{156} + \frac{1}{259} + \ldots \) In this series the fourth differences of the denominators = 0; therefore the general term must be represented by \( \frac{1}{ax^3 + bx^2 + cx + d} \); write therefore 2, 3, 4, &c. for \( x \), and we have \[ 8a + 4b + 2c + d = 15, \quad 27a + 9b + 3c + d = 40, \quad 64a + 16b + 4c + d = 85, \quad 125a + 25b + 5c + d = 156; \] hence \( a = 1, b = 1, c = 1, d = 1 \), and the general term is \( \frac{1}{x^3 + x^2 + x + 1} = \frac{1}{x^3} - \frac{1}{x^4} + \frac{1}{x^7} - \frac{1}{x^8} + \ldots \); hence the sum \( = B - C + F - G + K - L + \ldots = 1242700165 \). PROP. VI. To find the sum of \( \frac{2}{5^2} + \frac{3}{10^2} + \frac{4}{17^2} + \ldots \) ad inf. The general term \( = \frac{x}{x^2 + 1} = \frac{1}{x^3} - \frac{2}{x^5} - \frac{3}{x^7} + \ldots \); hence, by writing 2, 3, 4, &c. for \( x \), we have the sum \( = B - 2D + 3F - \ldots = 147115771469 \). In like manner \( \frac{2}{3^2} + \frac{3}{8^2} + \frac{4}{15^2} + \ldots = B + 2D + 3F + \ldots = 312498999865 \). PROP. VII. To find the sum of \( \frac{1}{3^2 \cdot 5^2} + \frac{1}{8^2 \cdot 10^2} + \frac{1}{15^2 \cdot 17^2} + \&c. \) ad inf. The general term is \( \frac{1}{x^2 - 1 \times x + 1} = \frac{1}{x^2} + \frac{2}{x^4} + \frac{3}{x^6} + \&c.; \) hence, by writing 2, 3, 4, &c. for \( x, \) we have the sum \( = G + 2L + 3P + \&c. = ,009447690684. \) PROP. VIII. To find the sum of \( \frac{1}{1^3 \cdot 2^3 \cdot 3^3} + \frac{1}{2^3 \cdot 3^3 \cdot 4^3} + \frac{1}{3^3 \cdot 4^3 \cdot 5^3} + \&c. \) ad inf. Here the general term is \( \frac{1}{x^3 - 1 \times x^3 \times x + 1} = \frac{1}{x^3} + \frac{3}{x^6} + \frac{6}{x^9} + \&c., \) and hence the sum \( = H + 3K + 6M + \&c. = ,004707148337. \) PROP. IX. To find the sum of the infinite series \( 1 - \frac{1}{3} + \frac{1}{6} - \frac{1}{10} + \&c. \) being a series of the reciprocal of the figurative numbers of the 3rd order, having the signs alternately + and −. This series, by resolving two terms into one, becomes \( \frac{4}{1 \cdot 2 \cdot 3} + \frac{4}{3 \cdot 4 \cdot 5} + \frac{4}{5 \cdot 6 \cdot 7} + \&c. \) whose general term, by writing 2, 4, 6, &c. for \( x, \) is \( \frac{4}{x^3 - 1 \times x \times x + 1} = \frac{4}{x^3} + \frac{4}{x^5} + \frac{4}{x^7} + \&c. \) consequently the sum \( = 4B'' + 4D'' + 4F'' + \&c. = \) (by Cor. Ex. 2. Prop. 2.) \( - 2 + 4 \) hyp. log. 2. Cor. Cor. Hence, as \(1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \&c. = 2\), we have \(1 + \frac{1}{6} + \frac{1}{15} + \&c. = 2\) hyp. log. 2, and \(\frac{1}{3} + \frac{1}{10} + \frac{1}{21} + \&c. = 2 - 2\) hyp. log. 2. **Prop. X.** To find the sum of the infinite series \(1 - \frac{1}{4} + \frac{1}{10} - \frac{1}{20} + \&c.\), being the reciprocals of the figurative numbers of the 4th order, having the signs alternately + and −. If we write 2, −3, 4, −5, &c. for \(x\), the general term will be \(\frac{6}{x^3 - x} = \frac{6}{x^3} + \frac{6}{x^5} + \&c.\); hence the sum required = \(6b + 6d + 6f + \&c. = (by \ Prop. 3.) - 7\frac{1}{2} + 12\) hyp. log. 2. Cor. Because the sum of \(1 + \frac{1}{4} + \frac{1}{10} + \frac{1}{20} + \&c. = \frac{3}{2}\); therefore \(1 + \frac{1}{10} + \frac{1}{35} + \&c. = -3 + 6\) hyp. log. 2; and \(\frac{1}{4} + \frac{1}{10} + \frac{1}{56} + \&c. = 4\frac{1}{2} - 6\) hyp. log. 2. **Prop. XI.** To find the sum of \(\frac{2^2}{1^2 \cdot 3^2} + \frac{3^2}{2^2 \cdot 4^2} + \frac{4^2}{3^2 \cdot 5^2} + \&c.\) ad infinitum. The general term, by writing 2, 3, 4, &c. for \(x\), is \(\frac{x^2}{x-1 \times x+1} = \frac{1}{x^2} + \frac{2}{x^4} + \frac{3}{x^6} + \&c.\); hence the sum = A + 2C + 3E + &c. = ,884966993407. **Prop.** PROP. XII. To find the sum of \( \frac{1}{1 \cdot 2^2 \cdot 3} + \frac{1}{2 \cdot 3^2 \cdot 4} + \frac{1}{3 \cdot 4^2 \cdot 5} + \&c. \) ad infinitum. Here the general term, by writing 2, 3, 4, &c. for \( x \), is \[ \frac{1}{x-1 \cdot x^2 \cdot x+1} = \frac{1}{x^6} - \frac{2}{x^7} + \frac{4}{x^8} + \frac{6}{x^9} - \frac{9}{x^{10}} - \frac{12}{x^{11}} + \frac{16}{x^{12}} - \&c.; \] consequently the sum \( = E - 2F + 4G - 6H + 9I - 12K + \&c. = 0.010370898482. \) PROP. XIII. To find the sum of \( \frac{1}{2}A - \frac{1}{4}B + \frac{1}{8}C + \&c. \) ad infinitum. The hyp. log. \( z = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \&c. = 1 + \frac{1}{3} + \frac{1}{5} + \&c. - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \&c.; \) hence \( 2 \times \text{hyp. log. } z, \text{ or hyp. log. } 4z, \) \[ = \frac{2}{1} + \frac{2}{3} + \frac{2}{5} + \&c. - 1 - \frac{1}{2} - \frac{1}{3} - \&c. \] Now, by division, \[ \frac{2}{2x+1} = \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{4x^3} - \frac{1}{8x^4} + \&c.; \] hence, by writing 2, 3, 4, &c. for \( x \), we have (after transposition) \( \frac{2}{5} + \frac{2}{7} + \&c. - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} - \&c. = - \frac{1}{2}A + \frac{1}{4}B - \frac{1}{8}C + \&c.; \) hence, by adding equal quantities to each side, we have \( \frac{2}{1} + \frac{2}{3} + \frac{2}{5} + \&c. - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} - \&c. = \frac{8}{3} - \frac{1}{2}A + \frac{1}{4}B - \frac{1}{8}C + \&c., \) consequently \( \frac{1}{2}A - \frac{1}{4}B + \frac{1}{8}C + \&c. = \frac{8}{3} - \text{hyp.log. } 4z. \) PROP. XIV. To find the sum of the infinite series \( \frac{1}{2 \cdot 5} + \frac{1}{3 \cdot 7} + \frac{1}{4 \cdot 9} + \&c. \) The general term, by writing 2, 3, 4, &c. for \( x \), is \[ \frac{1}{x \cdot 2x + 1} = \frac{1}{2x^2} - \frac{1}{4x^3} + \frac{1}{8x^4} - \&c.; \text{ hence the sum} = \frac{1}{2}A - \frac{1}{4}B + \frac{1}{8}C - \&c. = (\text{by Prop. 13.}) \frac{8}{3} - \text{hyp. log. } 4. \] PROP. XV. To find the sum of \( 1 + \frac{1}{2} + \frac{1}{3} + \&c. \) to \( \frac{1}{x} \). The hyp. log. \( \frac{x}{x-1} = \frac{1}{x} + \frac{1}{2x^2} + \frac{1}{3x^3} + \frac{1}{4x^4} + \&c.; \text{ consequently} \) hyp. log. \( \frac{x}{x-1} = \frac{1}{2x^2} - \frac{1}{3x^3} - \frac{1}{4x^4} - \&c. = \frac{1}{x}; \text{ hence, if we write} \) \( 2, 3, 4, \&c. \) for \( x \), we have hyp. log. \( \frac{2}{1} + \text{hyp. log. } \frac{3}{2} + \&c. \&c. \&c. \) \[ \begin{align*} \text{hyp. log. } \frac{x}{x-1} &= \frac{1}{2} \times \frac{1}{2^2} + \frac{1}{3^2} + \&c. \&c. \&c. \\ &= \frac{1}{3} \times \frac{1}{2^3} + \frac{1}{3^3} + \&c. \&c. \&c. \\ &= \frac{1}{4} \times \frac{1}{2^4} + \frac{1}{3^4} + \&c. \&c. \&c. \\ &= \&c. \&c. \&c. \end{align*} \] \( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \&c. \&c. \&c. \&c.; \text{ but hyp. log. } \frac{2}{1} + \text{hyp. log. } \frac{3}{2} + \&c. \&c. \&c. \&c.; \text{ hyp. log. } \frac{x}{x-1} = \text{hyp. log. } \frac{2}{1} \times \frac{3}{2} \times \frac{4}{3} \times \&c. \&c. \&c. \&c.; \text{ hyp. log. } x; \text{ also } \frac{1}{2^2} + \frac{1}{3^2} + \&c. \&c. \&c. \&c.; \text{ the sum} \] sum of the same series ad infinitum, minus the sum of all the terms from \( \frac{1}{x^2} = (if x + 1 = n) A - \frac{1}{n} - \frac{1}{2n^2} - \frac{1}{6n^3} + \frac{1}{30n^5} - \frac{1}{42n^7} + &c.; \) in the same manner \( \frac{1}{2^3} + \frac{1}{3^3} + &c. \ldots \frac{1}{x^3} = B - \frac{1}{2n^2} - \frac{1}{2n^3} - \frac{1}{4n^4} + \frac{1}{12n^5} - \frac{1}{12n^8} + &c.; \) and so on for the other series; hence, by substitution, and adding unity to each side, we have hyp. log. \( x + 1 - \frac{1}{2}A - \frac{1}{3}B - \frac{1}{4}C - &c. + \frac{1}{2n} + \frac{5}{12n^2} + \frac{1}{3n^3} + \frac{31}{120n^4} + \frac{1}{5n^5} + \frac{41}{252n^6} + \frac{61}{7n^7} + \frac{31}{240n^8} + &c. = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + &c. \ldots \frac{1}{x}; \) but \( 1 - \frac{1}{2}A - \frac{1}{3}B - \frac{1}{4}C - &c. = ,577215664901; \) hence \( 1 + \frac{1}{2} + \frac{1}{3} + &c. \ldots \frac{1}{x} = \text{hyp. log. } x + ,577215664901 + \frac{1}{2n} + \frac{5}{12n^2} + \frac{1}{3n^3} + \frac{31}{120n^4} + \frac{1}{5n^5} + \frac{41}{252n^6} + \frac{1}{7n^7} + \frac{31}{240n^8} + &c. \) Ex. 1. Let \( x = 10000; \) then \[ \begin{align*} \text{hyp. log. } 10000 &= 9,210340371976 \\ \text{const. quant.} &= ,577215664901 \\ \frac{1}{2n} &= ,000049995000 \\ \frac{5}{12n^2} &= ,000000004166 \\ \end{align*} \] therefore the sum required \( = 9,787606036043 \) Ex. 2. Let \( x = 10000000; \) then \[ \begin{align*} \text{hyp. log. } 10000000 &= 16,118095650958 \\ \text{const. quant.} &= ,577215664901 \\ \frac{1}{2n} &= ,000000049999 \\ \end{align*} \] therefore the sum required \( = 16,695311365858 \) PROP. XVI. To find the value of $\alpha \times \beta \times \gamma \times \delta \times \&c.$ ad infinitum, supposing the general term to be a rational function of $x$. Let $\pi$ be the general term, then resolve $\frac{\dot{x}}{\pi}$ into an infinite series, and take the fluent on both sides; then write 2, 3, 4, &c. for $x$, and one side will become the hyp. log. of the given series, and the value of the other side may be found from the tables. Ex. 1. To find the value of $\frac{4}{3} \times \frac{9}{8} \times \frac{16}{15} \times \&c.$ ad infinitum. Here the general term is $\frac{x^2}{x^2 - 1}$; hence $\frac{\dot{x}}{\pi} = -\frac{2\dot{x}}{x^3 - x} = -\frac{2\dot{x}}{x^3} - \frac{2\dot{x}}{x^7} - \&c.$; hence the hyp. log. $\frac{x^2}{x^2 - 1} = \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \&c.$ Write 2, 3, 4, &c. for $x$, and we have the hyp. log. $\frac{4}{3} + \text{hyp. log.} \frac{9}{8} + \text{hyp. log.} \frac{16}{15} + \&c. = A + \frac{1}{2}C + \frac{1}{3}E + \&c. = ,693147180574,$ which is the hyp. log. 2; but hyp. log. $\frac{4}{3} + \text{hyp. log.} \frac{9}{8} + \text{hyp. log.} \frac{16}{15} + \&c. = \text{hyp. log.} \frac{4}{3} \times \frac{9}{8} \times \frac{16}{15} \times \&c.$ consequently $\frac{4}{3} \times \frac{9}{8} \times \frac{16}{15} \times \&c. = 2.$ Ex. 2. To find the value of $\frac{8}{7} \times \frac{27}{26} \times \frac{64}{63} \times \&c.$ ad infinitum. Here the general term is $\frac{x^3}{x^3 - 1}$; hence $\frac{\dot{x}}{\pi} = -\frac{3\dot{x}}{x^4 - x} = -\frac{3\dot{x}}{x^4} - \frac{3\dot{x}}{x^7} - \frac{3\dot{x}}{x^{10}} - \&c.$; hence the hyp. log. $\frac{x^3}{x^3 - 1} = \frac{1}{x^3} + \frac{1}{2x^6} + \frac{1}{3x^9} + \&c.$ Write 2, 3, 4, &c. for $x$, and we have hyp. log. $\frac{8}{7} + \text{hyp. log.} \frac{27}{26}$ of Infinite Series. \[ \frac{27}{26} + \text{hyp. log. } \frac{64}{63} + \&c. = B + \frac{1}{2}E + \frac{1}{3}H + \&c. = ,211466250444; \] or hyp. log. \(\frac{8}{7} \times \frac{27}{26} \times \frac{64}{63} \times \&c. = ,211466250444\); hence \(\frac{8}{7} \times \frac{27}{26} \times \frac{64}{63} \times \&c. = 1,627295, \&c.\). Hence we may find the value of such a quantity, supposing the number of factors to be finite. Ex. To find the value of \(\frac{2}{1} \times \frac{4}{3} \times \frac{6}{5} \times \&c. \ldots\) to \(\frac{2x}{2x-1}\). Here the general term being \(\frac{2x}{2x-1}\), we have \(\frac{x}{\pi} = -\frac{\dot{x}}{2x^2} - \frac{\ddot{x}}{4x^3} - \frac{\dddot{x}}{8x^4} - \frac{\ddddot{x}}{16x^5} - \&c.;\) hence hyp. log. \(\frac{2x}{2x-1} = \frac{1}{1 \cdot 2x} + \frac{1}{2 \cdot 4x^2} + \frac{1}{3 \cdot 8x^3} + \frac{1}{4 \cdot 16x^4} + \&c.\). Now write \(2, 3, 4, \&c.\) for \(x\), and we have the hyp. log. \(\frac{4}{3} \times \frac{6}{5} \times \frac{8}{7} \times \&c. \ldots\) \[ \frac{2x}{2x-1} = \frac{1}{2} \times \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \frac{1}{x} \] \[ + \frac{1}{2 \cdot 4} \times \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \ldots \frac{1}{x^2} \] \[ + \frac{1}{3 \cdot 8} \times \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \ldots \frac{1}{x^3} \] \[ + \&c. \quad \&c. \quad \&c. \] But, by Prop. 15. \(\frac{1}{2} \times \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \frac{1}{x} = \frac{1}{2} \text{hyp. log.}\) \(x = ,211392167549 + \frac{1}{4n} + \frac{5}{24n^2} + \frac{1}{6n^3} + \frac{31}{240n^4} + \frac{1}{10n^5} + \frac{41}{504n^6} + \&c.;\) also \(\frac{1}{2 \cdot 4} \times \frac{1}{2^2} + \frac{1}{3^2} + \ldots \frac{1}{x^2} = \frac{1}{2 \cdot 4} \times A - \frac{1}{n} - \frac{1}{2n^2} - \frac{1}{6n^3} + \frac{1}{30n^5} - \&c.\) and \(\frac{1}{3 \cdot 8} \times \frac{1}{2^3} + \frac{1}{3^3} + \ldots \frac{1}{x^3} = \frac{1}{3 \cdot 8} \times B - \frac{1}{2n^2} - \frac{1}{2n^3} - \frac{1}{4n^4} + \frac{1}{12n^6} - \&c.,\) and so on for the other serieses: hence, by substitution, tion, hyp. log. $\frac{4}{3} \times \frac{6}{5} \times \frac{8}{7} \times \&c. \ldots \ldots \frac{2x}{2x-1} = \frac{1}{2}$ hyp. log. $x =$ ,12078223764 + $\frac{1}{8n} + \frac{1}{8n^2} + \frac{23}{192n^3} + \frac{7}{64n^4} + \frac{61}{640n^5} + \&c.$; consequently $\frac{4}{3} \times \frac{6}{5} \times \frac{8}{7} \times \&c. \ldots \ldots \frac{2x}{2x-1} =$ the natural number corresponding to the right hand side of the equation; hence $\frac{2}{1} \times \frac{4}{3} \times \frac{6}{5} \times \ldots \ldots \frac{2x}{2x-1} =$ twice that natural number. Ex. Let $x = 10000$; then $\frac{1}{2}$ hyp. log. $x = 4,605170185988$; const. quant. $= ,120782237640$ $\frac{1}{8n} = 4,484387948548$ $\frac{1}{8n^2} = ,000012498750$ $\frac{1}{8n^3} = ,000000001249$ $\frac{20000}{19999} = 177,2476$ the natural number corresponding to which hyp. log. is 88,6238, &c., consequently $\frac{2}{1} \times \frac{4}{3} \times \frac{6}{5} \times \ldots \ldots \frac{20000}{19999} = 177,2476$. If $x$ be a very large number, it may be sufficiently exact in most cases to take twice the natural number corresponding to the hyp. log. of $\frac{1}{2}$ hyp. log. $x =$ ,120782237640.