# Limit of $\sum_{n=m+1}^{\infty} \frac{m}{n\sqrt{n^2-m^2}},m\rightarrow\infty$

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

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

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