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So that's the question. But I am unable to find a counter example if it's false. Or how should I proceed to prove if it is true? I am confused please help. Many thanks.

Harry Potter
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    Hint: for large enough $n$, $a_n^2 < a_n$ –  Nov 16 '15 at 10:49
  • If $\sum a_n$ is convergent, then $a_n<1$ for large enough $n$, so $a_n^2<a_n$, so it converges by domination. – J.R. Nov 16 '15 at 10:50
  • Oh didn't see that. Thank you guys. – Harry Potter Nov 16 '15 at 10:51
  • Note: They specify that $a_n > 0$ because of alternating series like $\sum_{n = 1}^\infty \frac{(-1)^n}{\sqrt{n}}$. It converges, but if you square every term, you get the harmonic series, which does not converge. – Arthur Nov 16 '15 at 10:53
  • Oh yes that's true. Good point @Arthur – Harry Potter Nov 16 '15 at 10:55
  • Notice that since terms $a_{n}$ are positive it follows that $\sum_{i = 1}^{n}a_{i}^{2}\leq (\sum_{i = 1}^{n}a_{i})^{2}$. Since $\sum a_{n}$ converges it follows that partial sums of $\sum a_{n}^{2}$ are bounded therefore it is convergent. – Paramanand Singh Nov 16 '15 at 11:38

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If $\sum a_n$ converges, then $0<a_n<1$ when $n$ is large (divergent test), then $0 < a_n^2 <a_n$ ($a_n >0$ is used) and one can conclude by comparison test.