December 30th, 2018, 04:05 AM  #1 
Senior Member Joined: Jul 2011 Posts: 407 Thanks: 16  Series Sum
Evaluation of $$\sum^{\infty}_{k=1}\frac{1}{k\cdot 2^k}$$

December 30th, 2018, 04:46 AM  #2 
Senior Member Joined: Oct 2009 Posts: 911 Thanks: 354 
Take the derivative of $\sum_k \frac{x^k}{k}$

January 2nd, 2019, 09:55 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, 10:59 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 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, 11:19 AM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 833 Thanks: 113 Math Focus: Elementary Math 
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 04:52 PM. 
January 2nd, 2019, 11:53 AM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,345 Thanks: 985 Math Focus: Wibbly wobbly timeywimey stuff.  
January 2nd, 2019, 01:53 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,855 Thanks: 744 
$\displaystyle f(x)=\sum_{k=1}^\infty \frac{x^k}{k},\ f'(x)=\sum_{k=1}^\infty x^{k1}=\frac{1}{1x}$. Therefore $f(x)=\ln(1x)$ or $f(\frac{1}{2})=\ln(2)$.
Last edited by skipjack; January 2nd, 2019 at 04:41 PM. 
March 19th, 2019, 07:39 AM  #8 
Senior Member Joined: Jul 2011 Posts: 407 Thanks: 16 
Thanks mathman.


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