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 November 2nd, 2019, 01:43 AM #11 Global Moderator   Joined: Dec 2006 Posts: 21,106 Thanks: 2324 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. Thanks from tahirimanov19 and idontknow November 2nd, 2019, 04:44 AM   #12
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 Originally Posted by idontknow 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. November 2nd, 2019, 05:05 AM   #13
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 Originally Posted by RDKGames 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. November 2nd, 2019, 05:09 AM #14 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math Prove : $\displaystyle ctg(x)=\frac{1}{x} -2x\sum_{n=1}^{\infty} \frac{1}{(xn\pi)^{2}}$ November 2nd, 2019, 05:27 AM #15 Newbie   Joined: Jun 2016 From: UK Posts: 8 Thanks: 8 Math Focus: Anything but stats :) 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 $\mathrm{ctg}(x)$. Thanks from idontknow November 2nd, 2019, 07:24 AM #16 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math Seems like the book is incorrect .  November 2nd, 2019, 08:44 AM   #17
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Quote:
 Originally Posted by idontknow Prove : $\displaystyle ctg(x)=\frac{1}{x} -2x\sum_{n=1}^{\infty} \frac{1}{(xn\pi)^{2}}$
I think it is $$\cot(x) = \dfrac{1}{x} + 2x \cdot \sum\limits_{k=1}^\infty \dfrac{1}{x^2 - \pi^2 k^2}$$ November 2nd, 2019, 09:17 AM #18 Senior Member   Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle Before thanks, you should check it first... I am not sure this is the exact formula... November 2nd, 2019, 12:11 PM #19 Global Moderator   Joined: Dec 2006 Posts: 21,106 Thanks: 2324 It looks correct, tahirimanov19, provided, of course, that the denominators are all non-zero. Tags compute, infinite, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Real Analysis 4 October 21st, 2019 11:03 PM poopeyey2 Advanced Statistics 6 July 10th, 2017 01:12 PM poopeyey2 Advanced Statistics 7 July 3rd, 2017 11:42 AM Shamieh Calculus 9 November 16th, 2013 05:53 PM momo Number Theory 0 April 18th, 2009 11:09 AM

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