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: 753 Thanks: 261 
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: 7,623 Thanks: 2611 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: iPhone Posts: 442 Thanks: 68 
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: 2,093 Thanks: 853 Math Focus: Wibbly wobbly timeywimey stuff.  
January 2nd, 2019, 12:53 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675 
$\displaystyle f(x)=\sum_{k=1}^\infty \frac{x^k}{k},\ f'(x)=\sum_{k=1}^\infty x^{k1}=\frac{1}{1x}$. 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.


Tags 
series, sum 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Using series manipulations to find the sums of finite series  stevewilliams  Algebra  1  April 22nd, 2016 01:40 PM 
Is finding laurent series expansion of f at z_0 using geometric series convenient?  king.oslo  Complex Analysis  0  December 28th, 2014 06:50 AM 
In need of help disk, series test, taylor, and power series  g0bearmon  Real Analysis  2  May 22nd, 2012 12:10 PM 
Convergent series > series of geometric means converges  The Chaz  Real Analysis  11  February 7th, 2011 04:52 AM 
In need of help disk, series test, taylor, and power series  g0bearmon  Calculus  1  December 31st, 1969 04:00 PM 