Examing a limit of sum of square roots

Question

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.

Thanks!

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 $1$. — Alan, Apr 11 '17 at 10:22
Sorry, this explanation is undecipherable. As far as I know, $1\le1/{\sqrt {10}}+1/{\sqrt {11}}+1/{\sqrt {12}}$ is false, for instance. — , Apr 11 '17 at 10:26
Surely for $n=1$ the middle term is $\frac{1}{\sqrt{2}}$, which is less than 1? — acernine, Apr 11 '17 at 10:27
@YvesDaoust I don't think I understand which inequality is wrong - the lower or higher bound? — Alan, Apr 11 '17 at 10:28
Does this answer your question? Find $\lim_\limits{n\to \infty}\left({1\over \sqrt{n^2+1}}+{1\over \sqrt{n^2+2}}+\cdots+{1\over \sqrt{n^2+n}}\right)$ — Sil, Apr 13 '20 at 17:19

score 3 · Accepted Answer · answered Apr 11 '17 at 10:34

3

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

then

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

answered Apr 11 '17 at 10:34

Thanks four all your help! – Alan Apr 11 '17 at 10:36

Examing a limit of sum of square roots

1 Answers1