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Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$

LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\frac{n}{\sqrt {n+1}}+\frac{n}{\sqrt {n+2}}.......\frac{n}{\sqrt {2n}}] = \infty$

Let $a_n=\frac{n}{\sqrt{2n}}$ then $a_n = \frac{1}{\sqrt 2}\sqrt{n}$

hence $\lim_{n\to\infty}a_n = \infty$

I continue the problem therefore using Cauchy's first theorem on limits.

But the way I took $a_n$ is it correct?

Does it not make $a_1 = \frac{1}{\sqrt{2}}$

If my way of taking $a_n$ is wrong, please suggest me the right way. Thank you

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If $s_n =\sum_{k=0}^n \dfrac1{\sqrt{n+k}} $ then $s_n \ge \sum_{k=0}^n \dfrac1{\sqrt{2n}} =(n+1)\dfrac1{\sqrt{2n}} \gt\dfrac{n}{\sqrt{2n}} =\dfrac{\sqrt{n}}{\sqrt{2}} \to \infty $

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Hint:

For each $k=0,1,\dots,n$, one has $\;\frac 1{\sqrt{n+k}}\ge\frac1{\sqrt{2n}}$. How many such terms do you have? Deduce a lower bound for the sum of these terms.

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Here is an elementary proof that

$\dfrac{\sqrt{2}}{3n} \lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1) \lt \dfrac{2\sqrt{2}}{n} $.

$\begin{array}\\ \sqrt{m+1}-\sqrt{m} &=(\sqrt{m+1}-\sqrt{m})\dfrac{\sqrt{m+1}+\sqrt{m}}{\sqrt{m+1}+\sqrt{m}}\\ &=\dfrac{1}{\sqrt{m+1}+\sqrt{m}}\\ &>\dfrac{1}{2\sqrt{m+1}}\\ \text{and}\\ \sqrt{m+1}-\sqrt{m} &<\dfrac{1}{2\sqrt{m}}\\ \end{array} $

so $2(\sqrt{m+1}-\sqrt{m}) <\dfrac{1}{\sqrt{m}} < 2(\sqrt{m}-\sqrt{m-1}) $.

Therefore

$\begin{array}\\ s_n &=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}\\ &\gt\sum_{k=0}^n 2(\sqrt{n+k+1}-\sqrt{n+k})\\ &=2(\sqrt{2n+1}-\sqrt{n})\\ &=2\sqrt{n}(\sqrt{2+1/n}-1)\\ &=2\sqrt{n}(\sqrt{2}\sqrt{1+1/(2n)}-1)\\ &\gt2\sqrt{n}(\sqrt{2}(1+1/(6n))-1) \quad\text{since }\sqrt{1+x} > 1+x/3 \text{ for } 0 < x < 3\\ &\gt2\sqrt{n}(\sqrt{2}-1)+\dfrac{\sqrt{2}}{3\sqrt{n}}\\ \text{and}\\ s_n &=\sum_{k=0}^n \dfrac1{\sqrt{n+k}}\\ &\lt\sum_{k=0}^n 2(\sqrt{n+k}-\sqrt{n+k-1})\\ &=2(\sqrt{2n}-\sqrt{n-1})\\ &=2\sqrt{n}(\sqrt{2}-\sqrt{1-1/n})\\ &\lt2\sqrt{n}(\sqrt{2}-(1-1/n)) \quad\text{since }\sqrt{1-x} > 1-x \text{ for } 0 < x < 1\\ &=2\sqrt{n}(\sqrt{2}-1)+\dfrac{2\sqrt{2}}{\sqrt{n}}\\ \end{array} $

Therefore $\dfrac{\sqrt{2}}{3n} \lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1) \lt \dfrac{2\sqrt{2}}{n} $.

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You could have quite good approximations using generalized harmonic numbers $$S_n=\sum_{i=0}^n \frac 1 {\sqrt{n+i}}=H_{2 n}^{\left(\frac{1}{2}\right)}-H_{n-1}^{\left(\frac{1}{2}\right)}$$ Using the asymptotics $$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2}\left(\frac{1}{p}\right)^{1/2}-\frac{1}{24} \left(\frac{1}{p}\right)^{3/2}+O\left(\frac{1}{p^{7/2}}\right)$$ Using it twice and continuing with Taylor series $$S_n=2 \left(\sqrt{2}-1\right) \sqrt{n}+\frac{\left(2+\sqrt{2}\right)}{4 \sqrt n} +O\left(\frac{1}{n^{3/2}}\right)$$ Just trying for $n=100$, the value should be $\approx 8.369654$ while the above truncated series gives $\frac{3}{40} \left(267 \sqrt{2}-266\right) \approx 8.369627$

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  • $\begingroup$ Hi Claude, you can also use Euler-Maclaurin summation. $\endgroup$ – Yves Daoust Jun 27 at 6:43
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$$\frac1{\sqrt n}\sum_{k=0}^n\frac1{\sqrt{n+k}}=\frac1n\sum_{k=0}^n\frac1{\sqrt{1+\dfrac kn}}\to\int_0^1\frac{dx}{\sqrt{1+x}}$$ is a Cauchy sum which converges to $$2(\sqrt2-1).$$

Your sum is $\sqrt n$ times larger.

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