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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 -> http://www.wolframalpha.com/input/?i=sum_k%3D1^1000+k*1000%2F%28k^2%2B1000%29^22012-06-04
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    Possible duplicate of [Asymptotic behavior of $\sum\limits_{n=1}^{\infty} \frac{nx}{(n^2+x)^2}$ when $x\to\infty$](http://math.stackexchange.com/questions/474263/asymptotic-behavior-of-sum-limits-n-1-infty-fracnxn2x2-when-x)2015-10-01
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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

1 Answers 1

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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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    nice solution. Thanks! How did you think of it?2012-06-04
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    Can you explain more for me why you have the greater part?2012-06-04
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    @CHris Yes, the first integral's value is $\,1/2\,$ . Please do note that it is the integral *with respect to $k$*!2012-06-04
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    @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.2012-06-04
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    @DonAntonio: notice that. Thanks.2012-06-04
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    @Micah: a very good thing to keep in my toolbox :)2012-06-04
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    @LongMai: It's basically the same reasoning as the first half, just using an upper Riemann sum instead of a lower one. See the formulas here: http://en.wikipedia.org/wiki/Summation#Approximation_by_definite_integrals2012-06-04
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    @Micah: this tool is often missed and it's such an elementary and powerful tool especially when one wants to find the limit by Squeeze theorem!2012-06-04
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    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.2015-10-01
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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