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Find value of $$\lim_{n\rightarrow \infty}\sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}}$$

$\bf{My\; Try::}$ Let $$\lim_{n\rightarrow \infty} S = \sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}} $$

So $$\lim_{n\rightarrow \infty} \underbrace{\frac{1}{\sqrt{n^2+2n+1}}+\frac{1}{\sqrt{n^2+2n+1}}+.......+\frac{1}{\sqrt{n^2+2n+1}}}_{\bf{(2n+2)\; times}}\leq \lim_{n\rightarrow \infty} S \leq \lim_{n\rightarrow \infty} \underbrace{\frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+.......+\frac{1}{\sqrt{n^2}}}_{\bf{(2n+2)\; times}}$$

So $$\frac{2n+2}{\sqrt{n^2+2n+1}}\leq S \leq \lim_{n\rightarrow \infty} \frac{2n+2}{\sqrt{n^2}}$$

So Using Sqeeze Theorem, WE get $$\lim_{n\rightarrow \infty} S = \lim_{n\rightarrow \infty} \sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}} = 2$$

My question is can we solve it any other way, If yes then plz explain here

Thanks

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Another way: $$ \int_{n^2}^{(n+1)^2}{1\over\sqrt x}dx=2 $$ This works since the difference between the sum and the integral approaches zero as $n\to\infty$.

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Another way using harmonic numbers $$A_n=\sum^{(n+1)^2}_{k=n^2}\frac{1}{\sqrt{k}}=H_{n^2+2 n+1}^{\left(\frac{1}{2}\right)}-H_{n^2-1}^{\left(\frac{1}{2}\right)}$$ Now, using, for large values of $m$ $$H_{m}^{\left(\frac{1}{2}\right)}=2 \sqrt{m}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{m}}+O\left(\frac{1}{m^{3/2}}\right)$$ and using again expansions leads to $$A_n=2+\frac{1}{n}-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$

For $n=10$, the above approximation gives $2.09500$ while the exact computation of the sum would give $\approx 2.09546$.

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