let $s_n$ denote the $n^{th}$ partial sum of series$$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+....$$Show $s_n>\sqrt{n}$ whenever $n>1$. Hence explain why series diverges.

I get the last bit, it's probably because $\sqrt{n}$ diverges and since $s_n>\sqrt{n}$, the so does $\sqrt{n}$ by comparison. But I'm stuck on the first part. Please help.

  • $\begingroup$ I tried using integration and the area under curves but couldn't get it to work. The curve I was comparing it with was $1/\sqrt{k}$ $\endgroup$
    – user117498
    Nov 4, 2014 at 5:11
  • $\begingroup$ Try induction. Now once you have the result $s_n > \sqrt{n}$, do you see how to use that to show the series diverges? $\endgroup$
    – Simon S
    Nov 4, 2014 at 5:12

2 Answers 2



$$n=2\;:\;\;1+\frac1{\sqrt2}>\sqrt2\iff\frac32+\sqrt2>2\iff 2>\frac14\;\;\color{red}\checkmark$$

Assume for $\;n\;$ and prove for $\;n+1\;$ :

$$1+\frac1{\sqrt2}+\ldots+\frac1{\sqrt n}+\frac1{\sqrt{n+1}}\stackrel{\text{Ind. Hyp.}}>\sqrt n+\frac1{\sqrt{n+1}}$$

Thus, it is enough to prove

$$\sqrt n+\frac1{\sqrt{n+1}}>\sqrt{n+1}\iff n+\frac1{n+1}+2\sqrt\frac n{n+1}>n+1\iff$$

$$\frac{4n}{n+1}>\frac{n^2}{(n+1)^2}\iff 4>\frac n{n+1}\iff3n+4>0\;\;\color{red}{\checkmark\checkmark}$$



Since $\sqrt{n}<\sqrt{n}+\sqrt{n-1}$ for $n>1$, then we have \begin{align*} \frac{1}{\sqrt{n}}&>\frac{1}{\sqrt{n}+\sqrt{n-1}}=\frac{n-(n-1)}{\sqrt{n}+\sqrt{n-1}}=\sqrt{n}-\sqrt{n-1} \end{align*}


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .