absolute convergence of $\sum _{n=1}^{\infty} a_n$ imply? Question is :
If $\sum _{n=1}^{\infty} a_n$ is absolutely convergent  then which of the following is not true?


*

*$\sum_{m=n}^{\infty}a_m\rightarrow 0$ as $n\rightarrow \infty$

*$\sum_{n=1}^{\infty}a_n\sin n$ is convergent.

*$\sum_{n=1}^{\infty}e^{a_n}$ is divergent.

*$\sum_{n=1}^{\infty}a_n^2$ is divergent.


First thing I would like to concentrate on is third option (as it is easy :P)....
absolutely convergence of $\sum _{n=1}^{\infty} a_n$ imply $a_n\rightarrow 0$ i.e., $e^{a_n}\rightarrow 1$ i.e.,$\sum_{n=1}^{\infty}e^{a_n}$ is divergent.
I guess second option is most probably true.. 
It is for sure absolute convergence as $|a_n\sin n|\leq |a_n|$ for all $n$....  I could not give concrete argument for convergence.
I guess fourth option is false...
absolutely convergence of $\sum _{n=1}^{\infty} a_n$ imply $a_n\rightarrow 0$ i.e., after certain stage $|a_n|<1$ i.e., $|a_n^2|<|a_n|$ So, we would have convergence of $\sum_{n=1}^{\infty}a_n^2$.
I do not understand what is actual point of first option... 
Could some one confirm if this justification for second/third/fourth options is sufficient and help me to understand what first option is...
Thank you.
 A: $\newcommand{\ra}{\operatorname{\rightarrow}}$
I assume that the terms of your series are real numbers.
1) First notice that the absolute convergence implies that each "infinite tail" $\sum_{n=N}^{\infty} a_n$ is convergent.The condition $\sum_{m=N}^{\infty} a_m \rightarrow 0$ as $N \rightarrow \infty$ is precisely the assertion that the sequence of partial sums $\sum_{k=1}^n a_k $is a Cauchy sequence.  Since $\mathbb{R}$ is complete, this is equivalent to the convergence of the series.  Since absolutely convergent series are convergent, this does follow from absolute convergence.
Here the absolute convergence was used only to see that $\sum_{m=N}^{\infty} a_m$ is a real number.  If we interpreted the statement as $\sup_{k \geq N} |\sum_{n=N}^{N+k} a_n| \ra 0$ as $N \ra \infty$, then we do not need absolute convergence to deduce this.
2) The inequality $|a_n \sin n| \leq |a_n|$ shows that $\sum_n |a_n \sin n| \leq \sum_n |a_n| < \infty$, so the series $\sum_n a_n \sin n$ is absolutely convergence and thus convergent.
Simple convergence of the $a_n$'s is not enough.  E.g. one could take $a_n = \frac{\sin n}{n}$.  It is delicate to show that this series is convergent -- one needs something like Dirichlet's Test -- but it does converge.  Then $\sum_n a_n \sin n = 
\sum_n \frac{ \sin^2 n}{n}$, and this is divergent because e.g. $|\sin n| \geq \frac{1}{2}$ for at least $\frac{1}{3}$ of the $n$'s between $1$ and $N$.  
3) What you say is correct.  Notice that you used much less than absolute convergence of the series but only that $a_n \ra 0$.
4) As you say, absolute convergence implies $a_n \ra 0$, hence $a_n^2 \leq |a_n|$ for sufficiently large $n$, hence by comparison $\sum_n a_n^2$ is convergent.   
Simple convergence of the $a_n$'s is not enough: take $a_n = \frac{(-1)^n}{n^{\frac{1}{2}}}$.
A: The last one is not true since for large $n$, $|a_n| < 1$ so $a_n^2 < |a_n|$ for all such $n$.
The rest are true.  Any series that is convergent must have its terms go to zero.  This is true by the cauchy criterion.  Choose $\epsilon > 0$.  Pick $N$ so $m, n \ge N\implies
|s_m - s_n| < \epsilon$, where the $s_n$ are the partial sums.  Let $m = n+1$; we have $|a_{n+1}| < \epsilon$ for $n\ge N$.  
