Calculus Calculus Math Forum

 December 30th, 2018, 04:05 AM #1 Senior Member   Joined: Jul 2011 Posts: 407 Thanks: 16 Series Sum Evaluation of $$\sum^{\infty}_{k=1}\frac{1}{k\cdot 2^k}$$ Thanks from greg1313 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 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, 01:53 PM #7 Global Moderator   Joined: May 2007 Posts: 6,855 Thanks: 744 $\displaystyle f(x)=\sum_{k=1}^\infty \frac{x^k}{k},\ f'(x)=\sum_{k=1}^\infty x^{k-1}=\frac{1}{1-x}$. 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 04:41 PM. March 19th, 2019, 07:39 AM #8 Senior Member   Joined: Jul 2011 Posts: 407 Thanks: 16 Thanks mathman. Tags series, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post stevewilliams Algebra 1 April 22nd, 2016 02:40 PM king.oslo Complex Analysis 0 December 28th, 2014 07:50 AM g0bearmon Real Analysis 2 May 22nd, 2012 01:10 PM The Chaz Real Analysis 11 February 7th, 2011 05:52 AM g0bearmon Calculus 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      