Find $\lim\limits_{n\to +\infty}\big(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\big)$?

Question

Computing the limit: $$ \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}} + \dfrac{1}{\sqrt{n^2 + 2}} + \cdots + \dfrac{1}{\sqrt{n^2 + n}}\right).$$

I've tried taking:

We have $\lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}}\right) = \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 2}}\right) = \ldots = \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + n}}\right) = 0.$

Then, we obtain $$ \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}} + \dfrac{1}{\sqrt{n^2 + 2}} + \cdots + \dfrac{1}{\sqrt{n^2 + n}}\right)\\ = \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 1}}\right) + \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + 2}}\right) + \cdots + \lim\limits_{n\to +\infty}\left(\dfrac{1}{\sqrt{n^2 + n}}\right) = 0.$$

Any other approach to this is welcome. Thanks!

Answer 1

That is not valid. The $n$ is also a variable, so the sum becomes larger as it goes.

I will do it in this way that \begin{align*} \sum_{k=1}^{n}\dfrac{1}{\sqrt{n^{2}+k}}\leq\sum_{k=1}^{n}\dfrac{1}{\sqrt{n^{2}+1}}=\dfrac{n}{\sqrt{n^{2}+1}}, \end{align*} and \begin{align*} \sum_{k=1}^{n}\dfrac{1}{\sqrt{n^{2}+k}}\geq\sum_{k=1}^{n}\dfrac{1}{\sqrt{n^{2}+n}}=\dfrac{n}{\sqrt{n^{2}+n}}, \end{align*} now $n/\sqrt{n^{2}+1},n/\sqrt{n^{2}+n}\rightarrow 1$, Squeeze Theorem concludes.

Answer 2

$\frac {n}{\sqrt{n^2+n}}\le\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\le\frac {n}{\sqrt{n^2+1}}$

Taking the limit as $n$ goes to infinity.

$\lim_\limits{n\to\infty}\frac {n}{\sqrt{n^2+n}}\le\lim_\limits{n\to\infty}\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\le\lim_\limits{n\to\infty}\frac {n}{\sqrt{n^2+1}}$ $1\le\lim_\limits{n\to\infty}\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \cdots + \frac{1}{\sqrt{n^2 + n}}\le 1$

Answer 3

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

$$\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}>\sum_{k=1}^n\frac{1}{\sqrt{n^2+n+0.25}}=\sum_{k=1}^n\frac{1}{n+0.5}=\frac{n}{n+0.5}$$

As $\displaystyle \lim_{n\to\infty}\frac{n}{n+0.5}=1$, by squeezing principle, $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}=1$

Answer 4

You can take $$\frac {1} {\sqrt{n^2+1}} \leq \frac {1} {\sqrt{n^2+1}} $$ $$\frac {1} {\sqrt{n^2+2}} < \frac {1} {\sqrt{n^2+1}} $$ $$...$$ $$\frac {1} {\sqrt{n^2+n}} < \frac {1} {\sqrt{n^2+1}} $$ And then $$ \frac {1} {\sqrt{n^2+n}} < \frac {1} {\sqrt{n^2+1}} $$ $$\frac {1} {\sqrt{n^2+n}} <\frac {1} {\sqrt{n^2+2}} $$ $$... $$ $$\frac {1} {\sqrt{n^2+n}} \leq \frac {1} {\sqrt{n^2+n}} $$ After that we have $$\frac {n} {\sqrt{n^2+n}} <\frac {1} {\sqrt{n^2+1 }} +\frac {1} {\sqrt{n^2+2}} ...+\frac {1} {\sqrt{n^2+n}} <\frac {n} {\sqrt{n^2+1 }} $$ and you apply limit to it.

Answer 5

Let us consider the sequence $$a_{n}=\sum_{k=1}^{n}\frac{1}{\sqrt{1+k/n^{2}}}.$$ Then $$a_{n+1}-a_{n}=\sum_{k=1}^{n+1}\frac{1}{\sqrt{1+k/\left(n+1\right)^{2}}}-\sum_{k=1}^{n}\frac{1}{\sqrt{1+k/n^{2}}}$$ $$=\frac{1}{\sqrt{1+1/\left(n+1\right)}}+\sum_{k=1}^{n}\left(\frac{1}{\sqrt{1+k/\left(n+1\right)^{2}}}-\frac{1}{\sqrt{1+k/n^{2}}}\right)$$ and $$\sum_{k=1}^{n}\left(\frac{1}{\sqrt{1+k/\left(n+1\right)^{2}}}-\frac{1}{\sqrt{1+k/n^{2}}}\right)=\int_{0}^{1}\sum_{k=1}^{n}\frac{k}{\left(k+\left(n+x\right)^{2}\right)^{3/2}}dx\leq\frac{1}{n^{3}}\sum_{k=1}^{n}k=\frac{n\left(n+1\right)}{2n^{3}}\rightarrow0$$ as $n\rightarrow\infty$ so $$a_{n+1}-a_{n}=\frac{a_{n+1}-a_{n}}{n+1-n}\rightarrow1$$ then, by the Stolz-Cesàro theorem, $$\sum_{k=1}^{n}\frac{1}{\sqrt{n^{2}+k}}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+k/n^{2}}}\rightarrow \color{red}{1}.$$