Convergence of $\sum a_{n}$,$\sum a_{n}^{2}$ and $\sum a_{n}^{4}$ A Convergent series  of  real  numbers  $\sum a_{n}$  is  given , what  can  be  said  about  the  convergence  of  $\sum a_{n}^{2}$ and  $\sum a_{n}^{4}$.
Also , if  only  absolute  convergence  of $\sum a_{n}$ , i.e. convergence  of  $\sum |a_{n}|$  is  given  what  about  convergence  of  $\sum a_{n}^{2}$ and  $\sum a_{n}^{4}$. 
Please  give  me  some  hints as  to  how  to  proceed. 
Thanks.
 A: Hint:
Look at the two cases of $a_n=\frac{(-1)^n}{n^{1/4}}$ and $a_n=\frac{1}{n^2}$ for the first question. 
Now, for absolute convergence (or equivalently if $a_n$ is non-negative), use the sandwich theorem to show convergence of the (non-negative) series of general term $a_n^2$ and $a_n^4$. (For $n$ big enough, $\lvert a_n\rvert < 1$: this will be useful).
A: If all that you know is that $\sum a_n$ converges, nothing can be said about the convergence of $\sum a_n^2$ and $\sum a_n^4$.
Proexample: Let $a_n=0$. Here $\sum a_n$ converges, and $\sum a_n^2$ and $\sum a_n^4$ converge.
Counterexample: Let $a_n=(-1)^n\frac{1}{\sqrt[4]{n}}$.  Here $\sum a_n$ converges by the alternating series test, but neither $\sum a_n^2$ nor $\sum a_n^4$ converge ($p$-test).
If, however, you know that $\sum a_n$ converges absolutely, then $\sum a_n^2$ and $\sum a_n^4$ must converge absolutely as well, so a fortiori they must converge.
Proof: Since the series $\sum a_n$ converges, the sequence $a_n$ must also converge.  Thus there must exist an $N\geq 0$ such that $n \geq N$ implies $|a_n|<1$.  Since $\sum |a_n|$ converges, we can define $S=\sum_{n=N}^\infty |a_n|$.  Then, as $|a_n^2|\leq|a_n|$ and $|a_n^4|\leq|a_n|$ for $n\geq N$, we have $\sum_{n=0}^\infty |a_n^2|\leq\sum_{n=0}^{N-1}|a_n^2|+S$ and $\sum_{n=0}^\infty |a_n^4|\leq\sum_{n=0}^{N-1}|a_n^4|+S$, both of which are finite.  Thus $\sum a_n^2$ and $\sum a_n^4$ converge absolutely, which implies that they converge.
A: If $\sum a_n$ is convergent, then is $\sum a_n^2$, $\sum a_n^4$, or, for the matter of fact, $\sum a_n^k$ for any $k > 1$, convergent too. To see this, since $\sum a_n$ is convergent, then $|a_n| < 1$ for all $n > N$. Then $|a_n^k| < |a_n|$ if $k > 1$ and $$\sum a_n^k < \sum a_n,$$ thus $\sum a_n^k$ is convergent.
A: I know one result which says that product of two convergent series is convergent if atleast one of the two series converges absolutely.Use this result.
A: Consider the following property of a converging series:$$\lim_{n\to\infty}a_n=0$$Because if a series does converge, then $$\lim_{m\to\infty}\sum_{n=0}^{m}a_n=\lim_{m\to\infty}\sum_{n=0}^{m+1}a_n$$Due to the properties of $\infty$.  We then get $$\lim_{m\to\infty}a_{m+1}=0$$which is equivalent to my above statement.
Now consider $\sum a_n^2$.  $\lim_{n\to\infty}a_n^2=0$  In particular, it will approach $0$ faster than $a_n$.  So while $\sum a_n^2$ may produce values that are larger than $\sum a_n$, eventually, both summations will converge on some value.
If the first sum converges, then its squares will converge faster because $\lim_{n\to\infty}\frac{a_n^2}{a_n}=\lim_{n\to\infty}a_n=0$, meaning $a_n^2$ approaches $0$ faster.
However, the squares may not approach the same value as the original sum.
Also consider a hyper-cube/prism with n-dimensions.  You want to measure the longest diagonal that goes from one vertex to the opposite vertex.  Given side lengths $a_1,a_2,a_3,a_4,\cdots$, the solution would be $S=\sqrt{\sum a_n^2}$ where $S$ is the length of the longest diagonal.
In the same hypercube, we note $\sum a_n>S$.
This is all just manipulations of Pythagorean theorem for multiple side lengths along with the theory on triangles stating $$a+b>c$$where c is the longest side length.
I don't know anywhere $\sum a_n^4$ would be applied.
