My Math Forum Series Sum

 Calculus Calculus Math Forum

 December 30th, 2018, 03:05 AM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Series Sum Evaluation of $$\sum^{\infty}_{k=1}\frac{1}{k\cdot 2^k}$$ Thanks from greg1313
 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

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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by oldbrother K --> infinite 2^K--> infinite K*2^K --> infinite Therefore, the result is 0
That only means that the larger k is the smaller $\displaystyle \dfrac{x^k}{k}$ becomes. It says nothing about the sum.

-Dan

 January 2nd, 2019, 12:53 PM #7 Global Moderator   Joined: May 2007 Posts: 6,762 Thanks: 697 $\displaystyle f(x)=\sum_{k=1}^\infty \frac{x^k}{k},\ f'(x)=\sum_{k=1}^\infty x^{k-1}=\frac{1}{1-x}$. Therefore $f(x)=-\ln(1-x)$ or $f(\frac{1}{2})=\ln(2)$. Thanks from greg1313, topsquark and idontknow 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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