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I try to solve the following tricky limit:

$\lim_{x\rightarrow\infty} \sum_{k=1}^{\infty} \frac{kx}{(k^2+x)^2} $

For some large values, W|A shows that its limit tends to $\frac{1}{2}$ but not sure how to prove that.

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    What do you mean "for some large values"? Do you mean that the expression tends to $\frac 12$ as $k\to\infty$? That is not the same thing!2015-10-01

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ETA: These bounds are wrong, as $\frac{kx}{(k^2+x)^2}$ is not monotone in $k$. For a fixed version of this answer, see robjohn's answer here.


Notice that, for fixed $x$, your sum is less than $\int_0^\infty \frac{kx}{(k^2+x)^2} \, dk=\frac{1}{2}\, ,$ and greater than $\int_1^\infty \frac{kx}{(k^2+x)^2} \, dk=\frac{x}{2(1+x)} \, ,$ and then apply the squeeze theorem.

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    @robjohn: Good point. I think any attempt to fix this issue would essentially turn the answer into a copy of yours. So since it's accepted and I can't delete, I'll just stick a pointer there...2015-10-01