My Math Forum Compute infinite sum

 Real Analysis Real Analysis Math Forum

 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
Newbie

Joined: Jun 2016
From: UK

Posts: 8
Thanks: 8

Math Focus: Anything but stats :)

Quote:
 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
Senior Member

Joined: Dec 2015
From: Earth

Posts: 823
Thanks: 113

Math Focus: Elementary Math
Quote:
 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
Senior Member

Joined: Mar 2015
From: Universe 2.71828i3.14159

Posts: 169
Thanks: 65

Math Focus: Area of Circle
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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top