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