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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.

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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.
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  • $\begingroup$ 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? $\endgroup$ – ketan Jan 14 '15 at 7:18
  • $\begingroup$ plz answer above query $\endgroup$ – ketan Jan 14 '15 at 7:25
  • $\begingroup$ , coz if a_n is 1/n^2 then option 3 diverges $\endgroup$ – ketan Jan 14 '15 at 7:52
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    $\begingroup$ $a_n$ cannot be $1/n^2$, since $\sqrt[n]{1/n^2}\to1$. For your first question note that $\sqrt[n]{\sqrt{a_n}}=\sqrt{\sqrt[n]{a_n}}\to\sqrt{L}$ $\endgroup$ – sranthrop Jan 14 '15 at 8:04

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