If squared series is convergent then series is convergent absolutely? If $\sum_{i=0}^\infty a_n^2$ is convergent, does that imply $\sum_{i=0}^\infty a_n$ is convergent?
How can I prove that?
Thanks for the help.
 A: We have the counterexample $a_n=\frac1n$.  The series $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ converges.  But the harmonic series $\sum_{n=1}^\infty \frac1{n}$ is divergent.
A: As mentioned in other answers, $a_k=\frac1n$ is a counterexample, since $\sum_{k=1}^\infty \frac1{n}$ diverges and $\sum_{k=1}^\infty \frac1{n^2}$ converges. If you want to see proofs of these facts, you can look here:


*

*Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? and other questions linked there

*Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges and other questions linked there

*If you are also interested in the value of the sum $\sum_{k=1}^\infty \frac1{n^2}$: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ and other questions linked there
The implication in the opposite direction is true for $a_k>0$, i.e., for series with positive terms. See Prove that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ converges absolutely and other questions linked there.
A: Not true. Take $a_n=\dfrac{1}{n}$ for example. 
A: take $a_n=1/n$
then $\sum \frac{1}{n^2}$ is convergent while $\sum \frac{1}{n}$ is divergent
