Summing to a small value

Algebra Level 5

∑ n = 2 ∞ 1 ( n 2 − 1 ) ( n 2 ) ( n 2 + 1 ) \large \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{(n^2-1)(n^2)(n^2+1)}

If the above expression has a closed form a π b coth ⁡ ( π ) − π c d + e f \dfrac{a \pi}{b} \coth (\pi) - \dfrac{\pi^c}{d} + \dfrac{e}{f} , where a a , b b , c c , d d , e e and f f are positive integers with gcd ⁡ ( a , b ) = gcd ⁡ ( e , f ) = 1 \gcd(a,b) = \gcd(e,f) = 1 , find a + b + c + d + e + f a+b+c+d+e+f .


The answer is 28.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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2 solutions

Mark Hennings
Aug 8, 2017

We have S = ∑ n = 2 ∞ 1 ( n 2 − 1 ) n 2 ( n 2 + 1 ) = ∑ n − 2 ∞ 1 n 2 − 1 ( 1 n 2 − 1 n 2 + 1 ) = ∑ n = 2 ∞ [ 1 n 2 − 1 − 1 n 2 − 1 2 ( 1 n 2 − 1 − 1 n 2 + 1 ) ] = 1 2 ∑ n = 2 ∞ 1 n 2 − 1 − ∑ n = 2 ∞ 1 n 2 + 1 2 ∑ n = 2 ∞ 1 n 2 + 1 = 1 4 ∑ n = 2 ∞ ( 1 n − 1 − 1 n + 1 ) − ( ζ ( 2 ) − 1 ) + 1 2 ∑ n = 2 ∞ 1 n 2 + 1 = 1 4 ( 1 + 1 2 ) − ( 1 6 π 2 − 1 ) + 1 2 ∑ n = 2 ∞ 1 n 2 + 1 = 9 8 − 1 6 π 2 + 1 2 ∑ n = 1 ∞ 1 n 2 + 1 \begin{aligned} S & = \; \sum_{n=2}^\infty \frac{1}{(n^2-1)n^2(n^2+1)} \; = \; \sum_{n-2}^\infty \frac{1}{n^2-1}\left(\frac{1}{n^2} - \frac{1}{n^2+1}\right) \\ & = \; \sum_{n=2}^\infty \left[ \frac{1}{n^2-1} - \frac{1}{n^2} - \tfrac12\left(\frac{1}{n^2-1} - \frac{1}{n^2+1}\right)\right] \\ & = \; \tfrac12\sum_{n=2}^\infty \frac{1}{n^2-1} - \sum_{n=2}^\infty \frac{1}{n^2} + \tfrac12\sum_{n=2}^\infty \frac{1}{n^2+1} \\ & = \; \tfrac14\sum_{n=2}^\infty\left(\frac{1}{n-1} - \frac{1}{n+1}\right) - \big(\zeta(2) - 1\big) + \tfrac12\sum_{n=2}^\infty \frac{1}{n^2+1} \\ & = \; \tfrac14\big(1 + \tfrac12\big) - \big(\tfrac16\pi^2 - 1\big) + \tfrac12\sum_{n=2}^\infty \frac{1}{n^2+1} \\ & = \; \tfrac{9}{8} -\tfrac16\pi^2 + \tfrac12\sum_{n=1}^\infty \frac{1}{n^2+1} \end{aligned} Using the standard identity ∑ n = 1 ∞ 1 n 2 + 1 = 1 2 ( π coth ⁡ π − 1 ) \sum_{n=1}^\infty \frac{1}{n^2+1} \; = \; \tfrac12\big(\pi\coth\pi - 1\big) we deduce that S = 7 8 − 1 6 π 2 + 1 4 π coth ⁡ π S \; = \; \tfrac78 -\tfrac16\pi^2 + \tfrac14\pi\coth\pi making the answer 1 + 4 + 2 + 6 + 7 + 8 = 28 1 + 4 + 2 + 6 + 7 + 8 = \boxed{28} .

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@Mark Hennings - Do you know where I can find a proof of the "standard identity" you used? Thank you

Christopher Criscitiello - 3 years, 8 months ago

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You need a complex analysis result for an expansion of the cotangent function cot ⁡ z \cot z , using the fact that it is periodic of period π \pi and has simple poles of residue 1 1 at each integer. One treatment is given here .

Once you have that, put z = i z=i .

Mark Hennings - 3 years, 8 months ago

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@Mark Hennings - Nice source, thank you

Christopher Criscitiello - 3 years, 8 months ago

∑ n = 2 ∞ 1 ( n 2 − 1 ) n 2 ( n 2 + 1 ) = 1 2 ∗ ∑ n = 2 ∞ ( n 2 + 1 ) − ( n 2 − 1 ) ( n 2 − 1 ) n 2 ( n 2 + 1 ) = 1 2 ∗ ∑ n = 2 ∞ { 1 ( n 2 − 1 ) n 2 − 1 n 2 ( n 2 + 1 ) } = 1 2 ∗ ∑ n = 2 ∞ { n 2 − ( n 2 − 1 ) ( n 2 − 1 ) n 2 − ( n 2 + 1 ) − n 2 n 2 ( n 2 + 1 ) } = 1 2 ∗ ∑ n = 2 ∞ { 1 n − 1 − 1 n + 1 − 1 n 2 − 1 n 2 + 1 n 2 + 1 } = 1 2 ∗ ∑ n = 2 ∞ { 1 n − 1 − 1 n + 1 − 1 n 2 − 1 n 2 + 1 n 2 + 1 } = 1 2 ∗ ∑ n = 2 3 1 n − 1 + 1 2 ∗ ∑ n = 4 ∞ 1 n − 1 − ∑ n = 2 ∞ { 1 2 ∗ 1 n + 1 + 1 n 2 − 1 2 ( n 2 + 1 ) } = 1 2 ∗ ∑ n = 2 3 1 n − 1 + 1 2 ∗ ∑ n = 2 ∞ { 1 n + 1 − 1 n + 1 } − ∑ n = 2 ∞ { 1 n 2 − 1 2 ( n 2 + 1 ) } . . . . . . . B u t s e r i e s i s c o n v e r g i n g . = 1 2 ∗ ∑ n = 2 3 1 n − 1 − ∑ n = 2 ∞ { 1 n 2 − 1 2 ( n 2 + 1 ) } \begin{aligned} &~~~~~~~~~~~ \sum_{n=2}^\infty \dfrac1 {(n^2-1)n^2(n^2+1)} \\ &=\frac 1 2 *\sum_{n=2}^\infty \dfrac{(n^2+1)-(n^2-1)} {(n^2-1)n^2(n^2+1)} \\ &= \frac 1 2 *\sum_{n=2}^\infty \Big \{ \dfrac 1 {(n^2-1)n^2}- \dfrac 1 {n^2(n^2+1)} \Big \} \\ &=\frac 1 2 *\sum_{n=2}^\infty\Big \{ \dfrac {n^2-(n^2-1)} {(n^2-1)n^2}- \dfrac {(n^2+1)-n^2} {n^2(n^2+1)} \Big \} \\ &=\frac 1 2 *\sum_{n=2}^\infty\Big \{ \dfrac 1{ n-1}- \dfrac 1{ n+1}-\frac 1{ n^2} -\frac 1 { n^2}+ \dfrac 1 {n^2+1} \Big \} \\ &=\frac 1 2 *\sum_{n=2}^\infty\Big \{ \dfrac 1{ n-1}- \dfrac 1{ n+1}-\frac 1{ n^2} -\frac 1 { n^2}+ \dfrac 1 {n^2+1} \Big \} \\ &=\frac 1 2 *\sum_{n=2}^3 \dfrac 1{ n-1}+\frac 1 2 *\sum_{n=4}^\infty \dfrac 1{ n-1}-\sum_{n=2}^\infty\Big \{\frac 1 2* \dfrac 1 { n+1}+\frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} \\ &=\frac 1 2 *\sum_{n=2}^3 \dfrac 1{ n-1}+{\color{#3D99F6}{\ \frac 1 2 *\sum_{n=2}^\infty\Big \{ \dfrac 1{ n+1}- \dfrac 1{ n+1}\Big \}}} - \sum_{n=2}^\infty\Big \{ \frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} .......But~series~is ~converging.\\ &=\frac 1 2 *\sum_{n=2}^3 \dfrac 1{ n-1}-\sum_{n=2}^\infty\Big \{ \frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} \end{aligned} \\ W e h a v e ∑ n = 2 ∞ { 1 n 2 − 1 2 ( n 2 + 1 ) } = ∑ n = 1 ∞ { 1 n 2 − 1 2 ( n 2 + 1 ) } − ∑ n = 1 1 { 1 n 2 − 1 2 ( n 2 + 1 ) } = { π 2 6 − 1 4 ∗ π coth ⁡ π } − { 1 − 1 2 } ∴ ∑ n = 2 ∞ 1 ( n 2 − 1 ) n 2 ( n 2 + 1 ) = 1 2 ∗ ∑ n = 2 3 1 n − 1 − ∑ n = 2 ∞ { 1 n 2 − 1 2 ( n 2 + 1 ) } = 3 8 − { π 2 6 − 1 4 ∗ π coth ⁡ π − 1 2 } = 7 8 − π 2 6 + 1 4 ∗ π coth ⁡ π . = a b − π c d + e f ∗ π coth ⁡ π S o a + b + c + d + e + f = 1 + 4 + 2 + 6 + 7 + 8 = 28. \displaystyle We~ have~\sum_{n=2}^\infty\Big \{ \frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} \\ =\displaystyle \sum_{n=1}^\infty\Big \{ \frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} -\sum_{n=1}^1\Big \{ \frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} \\ =\displaystyle \Big \{ \dfrac {\pi^2} 6 - \tfrac 1 4*\pi\coth\pi \Big \} - \Big \{1 - \tfrac 1 2 \Big \}\\ \displaystyle \therefore~ ~~~~~~~~~~~ \sum_{n=2}^\infty \dfrac1 {(n^2-1)n^2(n^2+1)} \\ \displaystyle =\frac 1 2 *\sum_{n=2}^3 \dfrac 1{ n-1}-\sum_{n=2}^\infty\Big \{ \frac 1{ n^2} - \dfrac 1 {2(n^2+1)} \Big \} \\ \displaystyle =\frac 3 8 - \Big \{ \dfrac {\pi^2} 6 - \frac 1 4*\pi\coth\pi - \frac 1 2 \Big \} \\ \displaystyle =\frac 7 8 - \dfrac {\pi^2} 6 + \frac 1 4*\pi\coth\pi. \\ \displaystyle =\frac a b - \dfrac {\pi^c} d + \frac e f*\pi\coth\pi \\ \displaystyle So~~a+b+c+d+e+f=1+4+2+6+7+8= \Large \color{#D61F06}{28}.

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