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November 2nd, 2019, 01:43 AM   #11
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As $e^{-k}2^k = (2/e)^k$, put $x = 2/e$ in $\displaystyle \sum_{k=0}^\infty x^k = \frac{1}{1 - x}$.
Starting the sum at $k = 1$ subtracts 1 from the left-hand side, so subtract 1 from the right-hand side also.
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November 2nd, 2019, 04:44 AM   #12
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Can you post the method in short-terms ?
This is just an infinite geometric series with common ratio r=2e^(-1) <1 whose infinite sum has a nice formula that you can use.
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November 2nd, 2019, 05:05 AM   #13
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This is just an infinite geometric series with common ratio r=2e^(-1) <1 whose infinite sum has a nice formula that you can use.
Yes , my acknowledge. I have other summations.
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November 2nd, 2019, 05:09 AM   #14
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Prove : $\displaystyle ctg(x)=\frac{1}{x} -2x\sum_{n=1}^{\infty} \frac{1}{(xn\pi)^{2}}$
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November 2nd, 2019, 05:27 AM   #15
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Did you write that down correctly? $\mathrm{ctg}(x)$ is not equivalent to the right hand side.

Note that on the RHS we have that $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{(xn\pi)^2}$ is simply $\dfrac{1}{x^2\pi^2} \displaystyle \sum_{n=1}^{ \infty} \dfrac{1}{n^2}$ and $\displaystyle \sum_{n=1}^{ \infty} \dfrac{1}{n^2}$ is a standard/famous result yielding $\dfrac{\pi^2}{6}$.

Therefore your RHS reduces down to $\dfrac{1}{x} - \dfrac{2x}{x^2 \pi^2} \dfrac{\pi^2}{6}$ which is precisely $\dfrac{2}{3x}$ and not quite what we have for $\mathrm{ctg}(x)$.
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November 2nd, 2019, 07:24 AM   #16
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Seems like the book is incorrect .
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November 2nd, 2019, 08:44 AM   #17
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Prove : $\displaystyle ctg(x)=\frac{1}{x} -2x\sum_{n=1}^{\infty} \frac{1}{(xn\pi)^{2}}$
I think it is $$\cot(x) = \dfrac{1}{x} + 2x \cdot \sum\limits_{k=1}^\infty \dfrac{1}{x^2 - \pi^2 k^2}$$
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November 2nd, 2019, 09:17 AM   #18
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Before thanks, you should check it first... I am not sure this is the exact formula...
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November 2nd, 2019, 12:11 PM   #19
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It looks correct, tahirimanov19, provided, of course, that the denominators are all non-zero.
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