# Convergence of series $\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+…+\sqrt{n}}}$

$\sum_{n=1}^{\infty}{\frac{1}{\sqrt{1}+\sqrt{2}+...+\sqrt{n}}}$ $\\$

I tried a comparison test, Leibniz, Cauchy, Cauchy-integral and D'Alembert criteria but nothing gives me the result. Any hints would be helpful!

• First, find an asymptotic for $1+\sqrt2+\cdots+\sqrt n$. – Lord Shark the Unknown Jun 25 '17 at 12:48
• Do you need to find the value or to determine whether the sum converges? – Galc127 Jun 25 '17 at 12:51
• I think he wants to know whether the sum converges, but I'd really like to see if there is an explicit value! – Alberto Andrenucci Jun 25 '17 at 13:15
• @iskra Find a simple expression such that $f(n)\sim 1+\sqrt2+\cdots+\sqrt n$. – Lord Shark the Unknown Jun 25 '17 at 13:22
• @iskra why you don't find any help from Lord Shark's comment is not understandable. $1+\sqrt{2}+\dots +\sqrt{n}\thicksim cn^{3/2}$, $c>0$ do not depend on $n$. He is talking about $f(n)$ like $cn^{3/2}$. – MAN-MADE Jun 25 '17 at 17:19

## 3 Answers

hint

For $k=1,2,...n$,

$$\int_{k-1}^k\sqrt {t}dt <\sqrt {k}<\int_k^{k+1}\sqrt {t}dt$$

by sum,

$$\int_0^n\sqrt {t}dt <\sum_{k=1}^n\sqrt {k}.$$

Compare it with harmonic series. As $\sum_{k=1}^n\sqrt{k} = \Omega(n^{1.5})$, you can conclude that this series is convergent. And for the claim you can prove it by the following:

$$\sum_{k=1}^n\sqrt{k} \geq \sqrt{\frac{n}{2}} + \cdots+\sqrt{n}\geq \frac{n}{2}\sqrt{\frac{n}{2}}$$

Use AM-GM inequality and we get $\frac{\sqrt{1}+\sqrt{2}+\dots +\sqrt{n}}{n}>\sqrt[n]{(n!)^{\frac{1}{2}}}$ and then use the result $(n!)^2>n^n$, when $n=3,4,\dots$(Why?).

Note that $\sum_{n=1}^{\infty}\frac{1}{n^s}$ is finite if $s>1$

• lol! why I got downvoted? – MAN-MADE Jun 25 '17 at 13:17
• Minor quibble: the (strict) inequality $(n!)^2\gt n^n$ does not hold for $n=1$ or $2$. Otherwise very nice. (I saw the question a couple of hours ago and thought about using AM-GM, but I couldn't see an easy way to turn the factorial into an $n$th power.) – Barry Cipra Jun 25 '17 at 14:58
• @BarryCipra you are right, I should have mentioned it! But it is very known result, I thought that would not necessary – MAN-MADE Jun 25 '17 at 15:02
• If you change $\gt$ to $\ge$, then my minor quibble goes away. – Barry Cipra Jun 25 '17 at 15:22
• @BarryCipra more useful suggestion would be the edit I did now!! Since we are showing convergence, this minor edit will not effect much though, you may see my comment in this post, and may help to find better value math.stackexchange.com/questions/2335773/… – MAN-MADE Jun 25 '17 at 15:28