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How to evaluate $$\lim_{n \to \infty}\sum_{m=1}^{\infty}\int_{0}^{\infty} \left(\frac{ m+x}{(m^n+x^n)^n} \right )dx$$ I made the substitution $$x = mt$$and factored out $$m^{-(n^2-2)}$$. I got this: $$\lim_{n \to \infty} \left(\sum_{m=1}^{\infty}m^{-(n^{2}-2)}\right)\int_{0}^{\infty} (1+t)(1+t^n)^{-n} dt $$

After that I tried the substitution: $$t^{n}=tan^{2}\theta$$ but after substitution I got two beta integrals after which I couldn't proceed further. I got the following: $$\frac{2}{n}\lim_{n \to \infty} \left(\sum_{m=1}^{\infty}m^{-(n^{2}-2)}\right)\int_{0}^{\frac{π}{2}} (\sin^{(\frac{2}{n}-1)}{\theta}\cos^{(2n-\frac{2}{n}-1)}\theta + \sin^{(\frac{4}{n}-1)}{\theta}\cos^{(2n-\frac{4}{n}-1)}\theta) d\theta $$ I couldn't proceed further.The answer is 3/2. Could someone clarify ? Thank you.

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    $\begingroup$ It took me a while to type it in latex. Why -1 ? $\endgroup$ – user262378 Aug 1 '20 at 6:05
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By dominated convergence we can move the limit in. Then notice that only the $m=1$ term survives the summation and notice that

$$\lim_{n\to\infty} \frac{1}{(1+t^n)^n} = \begin{cases}0 & t \geq 1 \\ 1 & t < 1 \\ \end{cases}$$

so the limit becomes

$$\int_0^1 1+t\:dt = \frac{3}{2}$$

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  • $\begingroup$ I didn't know whether I could move the limit in. Thank you!! $\endgroup$ – user262378 Aug 1 '20 at 8:06

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