# 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?

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

• Do you know sandwich thm? – Vim May 24 '15 at 3:24

Each term is smaller than $\frac {1}{n \sqrt {n}}$. There are $n$ terms in the sum. Hence the sum is smaller than $\frac {1}{\sqrt {n}}$. In the limit of $n$ to $\infty$ the sum goes to $0$.

What about squeezing between $0$ and $\frac{n}{\sqrt{n^3+n^2}}$ which goes to zero?

• Rather $\frac n{\sqrt{n^3+1}}$. – Berci May 23 '15 at 23:54
• @Berci I don't understand. – Gregory Grant May 23 '15 at 23:55
• @GregoryGrant Berci is saying that the sum is bounded above by $\frac{n}{\sqrt{n^3 + 1}}$, since each summand is bounded above by $\frac{1}{\sqrt{n^3 + 1}}$. – kobe May 24 '15 at 0:07
• @kobe I see, but do we need that? – Gregory Grant May 24 '15 at 0:10
• @GregoryGrant it's to apply the squeeze theorem, as you've suggested. – kobe May 24 '15 at 0:28

Let, $$x_k=\frac{n}{\sqrt{n^3+k^2}}$$

Then, $$\lim_{n\to \infty}x_k=0.$$So, by Cauchy's first limit theorem

$$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^nx_k=0.$$

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 , $$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{\sqrt{n^3+k^2}}=0.$$