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I would like to show that $$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{2}$$

Using integrals:

$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} \leq \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}} \leq m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}}+\frac{m}{(m+1)\sqrt{2m+1}}$$

$$ m\int_{m+1}^{\infty} \frac{\mathrm dx}{x \sqrt{x^2-m^2}} = \frac{\pi}{2}-\arctan \left(1+\frac{1}{m} \right)=\frac{\pi}{4}+o(1)$$

The result I get is:

$$ \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}\rightarrow_{m\rightarrow \infty}\frac{\pi}{4}$$

Where did I go wrong?

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The integral is arcsec. You seem to have used $\arctan$. –  André Nicolas Jun 16 '12 at 13:31
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1 Answer

up vote 6 down vote accepted

The initial inequality appears to be set up correctly so the problem may lie with your evaluation of that integral. The result comes much quicker via Riemann Sums:

$$\sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}}= \frac{1}{m} \sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{(n/m)^2-1}} \to \int^{\infty}_1 \frac{1}{x\sqrt{x^2-1} } dx = \sec^{-1} x \bigg|^{\infty}_1= \frac{\pi}{2}.$$

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Thank you very much Ragib Zaman! –  Chon Jun 16 '12 at 14:15
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