Calculate the following limit: $$\lim\limits_{n \to\infty} \left(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}} \right)$$

I think this limit equals $1$. I am not sure. Tried using the squeeze theorem:

$$1 \leq \left(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}} \right) \leq n\cdot \frac{1}{\sqrt{n^2}}$$.

It's quite clear why the right hand side is bigger than the middle term, but is $1$ really smaller of equals to the middle term?

Please note I can't use integrals here nor taylor series.


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    $\begingroup$ Your left inequality is wrong. $\endgroup$ – Yves Daoust Apr 11 '17 at 10:19
  • $\begingroup$ @YvesDaoust thanks, can you please explain why? $\endgroup$ – Alan Apr 11 '17 at 10:20
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    $\begingroup$ Can you explain why you think it is right ? $\endgroup$ – Yves Daoust Apr 11 '17 at 10:20
  • $\begingroup$ Since the denominator in the right hand side is smaller than all the denominators in the middle term, and I multiply it $n$ times in the value of the nominator of the middle term $1$. $\endgroup$ – Alan Apr 11 '17 at 10:22
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    $\begingroup$ I said left and shown an example. Isn't that enough ? $\endgroup$ – Yves Daoust Apr 11 '17 at 10:28

For $1\le k\le n$,$$\frac1{(n+1)^2}<\frac1{n^2+k}<\frac1{n^2}$$


$$\frac n{n+1}<\sum_{k=1}^n\frac1{\sqrt{n^2+k}}<\frac nn.$$

  • $\begingroup$ Thanks four all your help! $\endgroup$ – Alan Apr 11 '17 at 10:36

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