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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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@Phira: i used W|A ->^1000+k*1000%2F%28k^2%2B1000%29^2 – user 1618033 Jun 4 '12 at 12:46
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! – Rory Daulton Oct 1 '15 at 20:13
up vote 16 down vote accepted

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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nice solution. Thanks! How did you think of it? – user 1618033 Jun 4 '12 at 12:49
Can you explain more for me why you have the greater part? – Long Mai Jun 4 '12 at 12:51
@CHris Yes, the first integral's value is $\,1/2\,$ . Please do note that it is the integral with respect to $k$! – DonAntonio Jun 4 '12 at 13:00
@Chris: In general, "try to approximate integrals with sums and vice versa" is a good thing to have in your toolbox. This sum looks particularly integral-like, since the $u$-substitution works out so nicely. – Micah Jun 4 '12 at 13:03
I know this question is a few years old, but I just answered a question which someone just pointed out was a duplicate of this one. Since the function being integrated is not monotonic, it is not so simple to tell how either integral compares to the sum. – robjohn Oct 1 '15 at 22:58

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