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October 3rd, 2017, 10:13 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: $\displaystyle \sum_1^\infty n^{2}e^{nx^2nx}$ (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, 01:28 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,434 Thanks: 562 
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, 02:03 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, 02:34 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,434 Thanks: 562 
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 $\displaystyle e^{x^2x}$ is an unnecessary complication. The problem is then $\displaystyle g(u)=\sum_{n=0}^\infty n^2u^n$, where $\displaystyle u=e^{x^2x}$. Starting at n=0 doesn't change anything. Step 1: $\displaystyle \sum_{n=0}^\infty n^2u^n=\sum_{n=0}^\infty n(n1)u^n+\sum_{n=0}^\infty nu^n=g_1(u)+g_2(u)$ Step 2: Look at $\displaystyle h(u)= \sum_{n=0}^\infty u^n=\frac{1}{1u}$, where u<1. Then $\displaystyle h'(u)=\sum_{n=0}^\infty nu^{n1}=\frac{1}{(1u)^2}$ and $\displaystyle h''(u)=\sum_{n=0}^\infty n(n1)u^{n2}=\frac{2}{(1u)^3}$ Put all this together to get $\displaystyle g_1=u^2h''(u)=\frac{2u^2}{(1u)^3}$ and $\displaystyle g_2=uh'(u)=\frac{u}{(1u)^2}$ so that $\displaystyle g(u)=\frac{u^2+u}{(1u)^3}$. 

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convergence, pointwise, power, sequence, sum, total 
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