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So how to approach this one? $\frac1n\sum g(\frac{r}{n}) $ . How to convert in this form? As I can see r and n will have different powers.

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    $\begingroup$ Where is $x$ in this some? $\endgroup$ – user64066 Mar 7 '15 at 13:49
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Since

$$\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + n}} + \cdots + \frac{1}{\sqrt{n^2 + n}} < \frac{1}{\sqrt{n^2 + 1}} + \cdots + \frac{1}{\sqrt{n^2 + n}}$$

and

$$\frac{1}{\sqrt{n^2 + 1}} + \cdots + \frac{1}{\sqrt{n^2 + n}} < \frac{1}{\sqrt{n^2 + 1}} + \cdots + \frac{1}{\sqrt{n^2 + 1}},$$

we have

$$\frac{n}{\sqrt{n^2 + n}} < \frac{1}{\sqrt{n^2 + 1}} + \cdots + \frac{1}{\sqrt{n^2 + n}} < \frac{n}{\sqrt{n^2 + 1}}.$$

As $$\frac{n}{\sqrt{n^2 + n}} = \frac{1}{\sqrt{1 + \frac{1}{n}}} \to 1$$

and

$$\frac{n}{\sqrt{n^2 + 1}} = \frac{1}{\sqrt{1 + \frac{1}{n^2}}} \to 1$$

by the squeeze theorem,

$$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\right) = 1.$$

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