December 30th, 2018, 03:05 AM  #1 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  Series Sum
Evaluation of $$\sum^{\infty}_{k=1}\frac{1}{k\cdot 2^k}$$

December 30th, 2018, 03:46 AM  #2 
Senior Member Joined: Oct 2009 Posts: 784 Thanks: 280 
Take the derivative of $\sum_k \frac{x^k}{k}$

January 2nd, 2019, 08:55 AM  #3 
Newbie Joined: Dec 2018 From: Canada Posts: 3 Thanks: 0 
K > infinite 2^K> infinite K*2^K > infinite Therefore, the result is 0 
January 2nd, 2019, 09:59 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,660 Thanks: 2635 Math Focus: Mainly analysis and algebra 
Yes, the terms must converge to zero for the series to converge (and they do, as you point out). No, the series doesn't converge to zero.

January 2nd, 2019, 10:19 AM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 514 Thanks: 80 
If not, how can we evaluate it? I know the method of derivatives, but it is not working. Last edited by skipjack; January 2nd, 2019 at 03:52 PM. 
January 2nd, 2019, 10:53 AM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,162 Thanks: 879 Math Focus: Wibbly wobbly timeywimey stuff.  
January 2nd, 2019, 12:53 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,762 Thanks: 697 
$\displaystyle f(x)=\sum_{k=1}^\infty \frac{x^k}{k},\ f'(x)=\sum_{k=1}^\infty x^{k1}=\frac{1}{1x}$. Therefore $f(x)=\ln(1x)$ or $f(\frac{1}{2})=\ln(2)$.
Last edited by skipjack; January 2nd, 2019 at 03:41 PM. 
March 19th, 2019, 06:39 AM  #8 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16 
Thanks mathman.


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