# $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n^2+0}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n^2+0}}+\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+3n}}+\cdots+\frac{1}{\sqrt{n^2+(n-1)n}}$

$\bf{My\; Solution::}$ We Can Write Given Series as

$$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=0}^{n-1} \frac{1}{\sqrt{n^2+r\cdot n}} = \lim_{n\rightarrow \infty}\sum_{r=0}^{n-1}\frac{1}{\sqrt{1+\frac{r}{n}}}\cdot \frac{1}{n}$$

Now Convert into Reinman Integral,

Put $\displaystyle \frac{r}{n} = x\;,$ Then $\displaystyle \frac{1}{n} = dx$ and Changing Limits, We get

$$\displaystyle \lim_{n\rightarrow \infty}\sum_{r=0}^{n-1}\frac{1}{\sqrt{1+\frac{r}{n}}}\cdot \frac{1}{n} = \int_{0}^{1}\frac{1}{\sqrt{1+x}}dx\;,$$

Put $(1+x) = t^2$ and $dx = 2tdt$ and Changing Limits, We Get

$$\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x}}dx =2 \int_{1}^{\sqrt{2}}\frac{t}{t}dt = 2\left(\sqrt{2}-1\right)$$

My Question is Can We solve It Using any Other Method Like

Using Squeeze Theorem or any Other Method.

Help Required Thanks,

• You may write all terms after limit in parenthesis (and +1) – L.G. Jul 25 '15 at 4:06
• (thread necromancy) Incidentally, I find $\int_{x=1}^2 \frac{dx}{\sqrt{x}}$ slightly less messy than $\int_{x=0}^1 \frac{dx}{\sqrt{1+x}}$. – Brian Tung Jul 26 '18 at 21:23

It can be shown that, for $\alpha\in(-1,0]$ and $N\in\mathbb{N}$, $$\frac{(N+1)^{\alpha+1}-1}{\alpha+1}\leq\sum_{k=1}^N\,k^\alpha\leq\frac{N^{\alpha+1}}{\alpha+1}\,.$$ Define $S_N$ for each $N\in\mathbb{N}$ to be $$\sum_{k=1}^N\,\frac{1}{\sqrt{k}}=\sum_{k=1}^N\,k^{-\frac{1}{2}}\,,$$ we have $$2(\sqrt{N+1}-1)\leq S_N \leq 2\sqrt{N}\,.$$
The required limit is $$\lim_{n\to\infty}\,\frac{S_{2n-1}-S_{n-1}}{\sqrt{n}}\,.$$ Now, for each $n\in\mathbb{N}$, $$2\left(\sqrt{2n}-1\right)-2\sqrt{n-1}\leq S_{2n-1}-S_{n-1} \leq 2\sqrt{2n-1}-2\left(\sqrt{n}-1\right)\,,$$ which means $$2\left(\frac{\sqrt{2n}-\sqrt{n-1}}{\sqrt{n}}\right)-\frac{2}{\sqrt{n}} \leq \frac{S_{2n-1}-S_{n-1}}{\sqrt{n}} \leq 2\left(\frac{\sqrt{2n-1}-\sqrt{n}}{\sqrt{n}}\right)+\frac{2}{\sqrt{n}}\,.$$ That is, $$\frac{2\left(n+1\right)}{\sqrt{n}\left(\sqrt{2n}+\sqrt{n-1}\right)}-\frac{2}{\sqrt{n}} \leq \frac{S_{2n-1}-S_{n-1}}{\sqrt{n}} \leq \frac{2\left(n-1\right)}{\sqrt{n}\left(\sqrt{2n-1}+\sqrt{n}\right)}+\frac{2}{\sqrt{n}}\,,$$ or $$\frac{2\left(1+\frac{1}{n}\right)}{\sqrt{2}+\sqrt{1-\frac{1}{n}}}-\frac{2}{\sqrt{n}} \leq \frac{S_{2n-1}-S_{n-1}}{\sqrt{n}} \leq \frac{2\left(1-\frac{1}{n}\right)}{\sqrt{2-\frac{1}{n}}+1}+\frac{2}{\sqrt{n}}\,.$$ Both sides of the inequalities above go to $\frac{2}{\sqrt{2}+1}=2\left(\sqrt{2}-1\right)$, as $n\to\infty$. By the Squeeze Theorem, $$\lim_{n\to\infty}\,\frac{S_{2n-1}-S_{n-1}}{\sqrt{n}}=2\left(\sqrt{2}-1\right)\,.$$
• You don't need the integral test to get the bounds of the sum, especially for $S_N$, in the first paragraph. You can use, for example, Bernoulli's Inequality. Since the OP wanted nothing to do with integrals, I chose a solution that doesn't involve any integral. – Batominovski Jul 25 '15 at 4:48