<s>[INST] Compute $\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^3+1}} + \frac{1}{\sqrt{n^3+4}} + \cdots + \frac{1}{\sqrt{n^3+n^2}}\right)$ How do I evaluate the following limit? 
I guess I should do a comparison, but I've got no clue about what to do. Could you give me a hand?

$$\lim_{n \to \infty}\left( \frac{1}{\sqrt{n^3+1}} + \frac{1}{\sqrt{n^3+4}}
 + \cdots + \frac{1}{\sqrt{n^3+n^2}}\right)$$

 [/INST] Easily we can find that, $$\frac{1}{\sqrt{n^3+n^2}}\le \frac{1}{\sqrt{n^3+k^2}}\le\frac{1}{\sqrt{n^3}}.$$for all $k=1,2,...,n$.
Taking summation, $$\sum_{k=1}^n\frac{1}{\sqrt{n^3+n^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3}}$$
$$\implies \frac{n}{\sqrt{n^3+n^2}}\le \sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}\le \frac{n}{\sqrt{n^3}}$$
Then , by Sandwich theorem , $$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}=0.$$
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