My Math Forum Pointwise, total convergence, and sum of a (power?) sequence
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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

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