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 lefthand side, so subtract 1 from the righthand side also. 
November 2nd, 2019, 04:44 AM  #12 
Newbie Joined: Jun 2016 From: UK Posts: 8 Thanks: 8 Math Focus: Anything but stats :)  
November 2nd, 2019, 05:05 AM  #13 
Senior Member Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math  
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 $\mathrm{ctg}(x)$. 
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 
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 Math Focus: Area of Circle  
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 nonzero.


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