$\sum a_{2n} $ converges Prove that if $\sum a_n$ converges absolutely, then $\sum a_{2n} $ converges.
I know this part, How it can be done but i am having problem in proving the later part of the question
T
o show that this result does not hold in general if we only assume that $\sum a_n$ converges conditionally. 
How am I suppose to do this question ?
 A: If $\sum a_n$ converges absolutely, then $b_{n}=a_{n}$ if $n$ is even and $b_{n}=0$ if $n$ is odd is a sequence which in absolute value is bounded above by $|a_n|$.  Since the series of $|a_n|$ converges, by direct comparison so does the second.  Note that the second ($b_n$'s) can be viewed as the same sequence as $a_{2n}$ in the sense that $a_{2n}$ selects the even terms out of the sequence $a_{n}$ and so does $b_{n}$.
For a counter-example (provided the sequence conditionally converges)... think of your favorite conditionally convergent sequence, the alternating harmonic series.
If you'd rather not have to show that the series containing only the positive terms of the alternating harmonic series diverges, you could use something more fun like the following:
$$ \frac{1}{1} - \frac{1}{1} + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \frac{1}{4} - \frac{1}{4} + \frac{1}{5}-\frac{1}{5}+...$$
If you only take every other term, you have the harmonic series (diverges).  But if you take all the terms, then the series converges to zero (values in the sequence of partial sums are either $\frac{1}{n}$ or 0 depending on how many terms you've summed).
A: If $\sum{a_n}$ absolutely converges, by Cauchy's criterion, $|a_n|+|a_{n+1}|+\ldots+|a_m|<\varepsilon$ for all $n,m>N$ for some $N\in\mathbb{N}$
Then, for all $n,m>\left \lfloor{\frac{N}{2}}\right \rfloor+1$, $|a_{2n}|+\ldots+|a_{2m}|<\varepsilon$ for the same fixed $\varepsilon.$ By Cauchy's criterion, it must converge.
Now, it is enough to put a counterexample to prove that being conditionally convergent may allow the result not to hold.
Choosing the alternating harmonic series, we can see that it converges to $\ln(2)$. However $\sum{a_{2n}}$ obviously does not converge.
