Hello I was wondering where I could start with computing this limit $\sum_{n=3}^\infty \frac{n^2}{2^n}$ . This is a proof based intro analysis class and recently we were taught epsilon neighborhoods but in this situation solving for $n$ doesn't seem like it will work. Any ideas of other ways to go about this ?
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1I don't understand what you mean – VIVID Sep 03 '20 at 17:30
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Do you want $\sum_{n=3}^\infty n^2/2^n$? or $\sum_{n=3}^\infty n^2/2^n$? – Angina Seng Sep 03 '20 at 17:31
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i think you have missed a sigma – Albus Dumbledore Sep 03 '20 at 17:32
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yes i forgot the sigma sorry ill figure out how to add that... dont usually use latex – bakes Sep 03 '20 at 17:32
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1Hint: $\sum n(n+1)x^{n-1}$ is an $f''(x)$. – zwim Sep 03 '20 at 17:39
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Are you trying to find the limit or show that it exists? Two very different questions. – pancini Sep 03 '20 at 17:40
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the question states to compute the limit so i think i need to actually find it – bakes Sep 03 '20 at 17:41
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Wolfram alpha can do it, if you want to check your answer. https://www.wolframalpha.com/input/?i=sum+n%3D3+to+infinity+n%5E2%2F2%5En – Stephen Montgomery-Smith Sep 03 '20 at 18:17
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We know that for $|x|<1$
$\displaystyle\sum_{n=0}^{+\infty} x^n=\frac{1}{1-x}$
Differentiating wrt $x$ and multiplying by $x$
$\displaystyle\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2} = x\sum_{n=1}^{\infty} nx^{n-1}$
By doing it another time, $\displaystyle\sum_{n=1}^{\infty} n^2x^n = \frac{x(1+x)}{(1-x)^3}$
I let you finish...
Bastien Tourand
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