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

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

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

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.

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

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 $$\frac{3}{40} \left(267 \sqrt{2}-266\right) \approx 8.369627$$

• Hi Claude, you can also use Euler-Maclaurin summation. – Yves Daoust Jun 27 at 6:43

$$\frac1{\sqrt n}\sum_{k=0}^n\frac1{\sqrt{n+k}}=\frac1n\sum_{k=0}^n\frac1{\sqrt{1+\dfrac kn}}\to\int_0^1\frac{dx}{\sqrt{1+x}}$$ is a Cauchy sum which converges to $$2(\sqrt2-1).$$

Your sum is $$\sqrt n$$ times larger.