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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?

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

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  • $\begingroup$ Do you know sandwich thm? $\endgroup$ – Vim May 24 '15 at 3:24
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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 $0$.

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

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

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

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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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