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 October 3rd, 2017, 09:13 AM #1 Newbie   Joined: Oct 2017 From: UK Posts: 2 Thanks: 0 Pointwise, total convergence, and sum of a (power?) sequence Hi guys... I seriously cannot help myself solving this problem: $\displaystyle \sum_1^\infty n^{2}e^{-nx^2-nx}$ (I'm not that good writing formulas but "n=1" and "+inf" are the sub and the sup) So I was asked to evaluate the pointwise and the total convergence of this sequence and find the sum. I have been thinking about to reduce it to a power sequence the fact is that I truly cannot think a way to do it.. Can you help me by resolving it and writing the steps? Thanks a lot! Best regards, An engineering student. October 3rd, 2017, 12:28 PM #2 Global Moderator   Joined: May 2007 Posts: 6,805 Thanks: 716 I am not sure what you mean by evaluate. For x=0, divergent. Otherwise convergent. Power series doesn't exist (as far as I can tell). October 3rd, 2017, 01:03 PM #3 Newbie   Joined: Oct 2017 From: UK Posts: 2 Thanks: 0 Yes, I probably make some mistakes while writing the first task of the exercise, ... maybe I got some problems with the translation since I am not English, sorry... but by the way I solved the first task.. what I cannot do now is to calculate the sum, that is the most difficult part...any ideas sir ? October 5th, 2017, 01:34 PM #4 Global Moderator   Joined: May 2007 Posts: 6,805 Thanks: 716 Good news! Sum is relatively easy. First note - divergence is for x in the closed interval from -1 to 0, not just 0. Sum: Recognize that $\displaystyle e^{-x^2-x}$ is an unnecessary complication. The problem is then $\displaystyle g(u)=\sum_{n=0}^\infty n^2u^n$, where $\displaystyle u=e^{-x^2-x}$. Starting at n=0 doesn't change anything. Step 1: $\displaystyle \sum_{n=0}^\infty n^2u^n=\sum_{n=0}^\infty n(n-1)u^n+\sum_{n=0}^\infty nu^n=g_1(u)+g_2(u)$ Step 2: Look at $\displaystyle h(u)= \sum_{n=0}^\infty u^n=\frac{1}{1-u}$, where |u|<1. Then $\displaystyle h'(u)=\sum_{n=0}^\infty nu^{n-1}=\frac{1}{(1-u)^2}$ and $\displaystyle h''(u)=\sum_{n=0}^\infty n(n-1)u^{n-2}=\frac{2}{(1-u)^3}$ Put all this together to get $\displaystyle g_1=u^2h''(u)=\frac{2u^2}{(1-u)^3}$ and $\displaystyle g_2=uh'(u)=\frac{u}{(1-u)^2}$ so that $\displaystyle g(u)=\frac{u^2+u}{(1-u)^3}$. Thanks from greg1313 Tags convergence, pointwise, power, sequence, sum, total 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 John Travolski Calculus 3 November 29th, 2016 07:09 PM Chemist@ Calculus 7 November 1st, 2014 02:16 PM Sandra93 Real Analysis 1 May 14th, 2014 06:12 AM rkose Number Theory 5 January 29th, 2010 07:08 AM DinkyDoeDoe Real Analysis 0 September 4th, 2009 12:33 PM

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