# To show series convergent if $\sum a_n$ converges

If$a_n> 0$ for n $\ge$1 and $\lim_{n\to \infty} {a_n}^{\frac{1}{n}}=L<1$ which of following series is convergent:

1. $\sum \sqrt{a_na_{n+1}}$
2. $\sum a_n^2$
3. $\sum \sqrt{a_n}$

4 $\sum \frac{1}{\sqrt{a_n}}$

I tried showing $\sum a_n^2$ convergent by comparison test, but I am unsure of how to show others.

## 1 Answer

Use the root test.

1. $\sqrt[n]{\sqrt{a_{n+1}a_n}}\to L<1$, convergent.
2. $\sqrt[n]{a_n^2}\to L^2<1$, convergent.
3. $\sqrt[n]{\sqrt{a_n}}\to \sqrt{L}<1$, convergent.
4. $\sqrt[n]{\frac{1}{\sqrt{a_n}}}\to\frac{1}{\sqrt{L}}>1$, divergent.
• How r u so sure, for example $\lim \sqrt{a_n}^{\frac{1}{n}} is \sqrt{L}$ is it some result from theorms of sequence? – ketan Jan 14 '15 at 7:18
• plz answer above query – ketan Jan 14 '15 at 7:25
• , coz if a_n is 1/n^2 then option 3 diverges – ketan Jan 14 '15 at 7:52
• $a_n$ cannot be $1/n^2$, since $\sqrt[n]{1/n^2}\to1$. For your first question note that $\sqrt[n]{\sqrt{a_n}}=\sqrt{\sqrt[n]{a_n}}\to\sqrt{L}$ – sranthrop Jan 14 '15 at 8:04