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It is difficult, so I need how to start to prove it $$\pi \tanh(\frac{\pi }{2})=\sum_{n=0}^{\infty }\frac{1}{n^2+n+0.5}$$

E.H.E
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    Do you know/can you use the partial fraction decomposition of $\pi \cot \pi z$? – Daniel Fischer Nov 03 '14 at 12:13
  • What is the problem in this question ????? – E.H.E Nov 03 '14 at 21:12
  • "What is the problem in this question" Quote: "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did Nov 05 '14 at 17:09

1 Answers1

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I am just mimicking the answer of @robjohn for this question, all credits go to him

let $x^2 + x + 0.5 = (x-a)(x-b)$, then we have $a+b = -1$, i.e. $b = -1-a$

\begin{align} &\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} \\ = &\sum_{n=0}^{+\infty}\dfrac{1}{(n-a)(n-b)}\\ = & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} - \dfrac{1}{n-b}\right)\\ = & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} - \dfrac{1}{n+ 1 + a}\right)\\ = & \dfrac{1}{a-b}\sum_{n=0}^{+\infty}\left(\dfrac{1}{n-a} + \dfrac{1}{-n- 1 - a}\right)\\ = & \dfrac{1}{a-b}\sum_{n=-\infty}^{+\infty}\dfrac{1}{n-a}\\ \end{align}

And we use the fact $\sum_{n=-\infty}^\infty\frac{1}{n-a}=-\pi\cot(\pi a)$,

so we have $$\sum_{n=0}^{+\infty}\dfrac{1}{n^2 + n + 0.5} = \dfrac{1}{b-a}\pi\cot(\pi a)$$

Now plug in the value of $a$ and $b$ to conclude.

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Petite Etincelle
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  • you talk about the trigonometric functions but our question is about the hyperbolic functions – E.H.E Nov 03 '14 at 12:23
  • @Ehegh $a$ and $b$ are complex numbers, plug their values and uses the complex definition of trigonometric functions, you can identify with your wanted result. – Petite Etincelle Nov 03 '14 at 12:28