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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 -> wolframalpha.com/input/?i=sum_k%3D1^1000+k*1000%2F%28k^2%2B1000%29^2 –  Chris's sis the artist Jun 4 '12 at 12:46

1 Answer 1

up vote 15 down vote accepted

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? –  Chris's sis the artist Jun 4 '12 at 12:49
Can you explain more for me why you have the greater part? –  Long 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
@DonAntonio: notice that. Thanks. –  Chris's sis the artist Jun 4 '12 at 13:07

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