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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,

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

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  • $\begingroup$ Well Done!! A + 1. Just a couple of suggestions. You might mention the use of the result of the integral test to establish the initial bounds for the sum. And you might also mention the straightforward algebra that established the limit after the sentence "The required limit is." $\endgroup$ – Mark Viola Jul 25 '15 at 4:39
  • $\begingroup$ 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. $\endgroup$ – Batominovski Jul 25 '15 at 4:48
  • $\begingroup$ My suggestion was to consider stating the source of the inequality rather than stating "it can be shown." The entire development depends on this inequality and that makes it worth mentioning. $\endgroup$ – Mark Viola Jul 25 '15 at 4:51

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