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October 3rd, 2017, 09:13 AM   #1
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Unhappy 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.
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October 3rd, 2017, 12:28 PM   #2
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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).
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October 3rd, 2017, 01:03 PM   #3
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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 ?
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October 5th, 2017, 01:34 PM   #4
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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
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