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Finding value of $$\lim_{n\rightarrow \infty}\lim_{m\rightarrow \infty}\sum^{n}_{r=1}\sum^{mr}_{k=1}\frac{m^2n^2}{(m^2n^2+k^2)(n^2+r^2)}$$

what i try

$$\lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\sum^{mr}_{k=1}\frac{m^2n^2}{m^2n^2+k^2}\cdot \frac{1}{n}\sum^{n}_{r=1}\frac{n}{n^2+r^2}$$

$$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\frac{n}{n^2+r^2}=\int^{1}_{0}\frac{1}{1+x^2}dx = \frac{\pi}{4}$$ How do i solve first summation

help me please

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    $\begingroup$ What you have there does not seem right. The second sum in the problem definition depends on $r$ which is the index of the first sum. In the second equation you moved the outer sum into the inner, but then $r$ in $mr$ would not be defined. $\endgroup$
    – geguze
    Sep 10, 2019 at 17:14
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    $\begingroup$ geguze is correct. You cannot exchange the order the summations without removing $r$ from the upper limit. $r$ is a dummy variable in the problem, but suddenly becomes explicit in your version. Another problem, which may or may not apply here, is that it is not always possible to exchange the order of the limits. either. Though it may be useful to simply assume for now that they can be exchanged, in order to find the limit, you will eventually have to justify that exchange before you can be sure the result is correct. $\endgroup$ Sep 10, 2019 at 23:32

1 Answer 1

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You don't need to exchange the order of the limits or the summations to figure this out. Because the summation to $n$ is finite for each value of $n$, you can exchange the limit with respect to $m$ with the first summation, and just look at $$\lim_{m\to \infty}\sum^{mr}_{k=1}\frac{m^2n^2}{(m^2n^2+k^2)(n^2+r^2)}$$ Factoring out the $m^2$, the fraction becomes $$\frac{n^2}{\left(n^2+\dfrac{k^2}{m^2}\right)(n^2+r^2)}$$

Because $k \le mr$, the maximum value of the denominator, and therefore the minimum value of the fraction, occurs when $k = mr$. Therefore $$\sum^{mr}_{k=1}\frac{m^2n^2}{(m^2n^2+k^2)(n^2+r^2)}\ge \sum^{mr}_{k=1}\frac{n^2}{(n^2+r^2)(n^2+r^2)} = mr\frac{n^2}{(n^2+r^2)^2}$$ Taking the limit of the rh expression as $m \to \infty$ shows that everything rises without bound.

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