Convergence test I am given that $\sum\limits_{n=1}^\infty a_n$ is convergent. 
I need to determine whether $\sum\limits_{n=1}^\infty (a_n)^\frac{1}{3}\;$ and $\;\sum\limits_{n=1}^\infty (a_n)^2\;$ are also convergent.
Imagine that $a_n = \dfrac{1}{n^4}.\;$ I believe that this is convergent because it's converging to $0$.
Following  the same thought, if $\displaystyle a_n = \left(\frac{1}{n^4}\right)^2,\;$ it's convergent because it's converging to $0$.
Am I doing this correctly or there is some other way to prove this?
 A: Task: You need to establish whether the statements are true for all $(a_n)$. 


*

*To prove such a statement is true, you need to show it is always holds for any convergent $a_n$ (not just that it holds for some $a_n$). 

*But to conclude that a given statement is false, you can simply find a single $a_n$ that serves as a counterexample.
Clarification: Recall that IF a series $\sum a_n$ converges, THEN $\lim a_n \to 0.\quad(1)$ 
The converse of (1) does not hold: if $\lim_{n\to \infty} b_n = 0$, it doesn't necessarily follow that $b_n$ converges.

You can use the p-series test to find an $a_n$ such that$\sum_{n=1}^\infty a_n$ converges, but such that $\sum_{n=1}^\infty (a_n)^{1/3}$ diverges.
p-series test: Recall that for $a_n = \dfrac{1}{n^p}, \;$
$\displaystyle \sum_{n=1}^\infty \frac{1}{n^p}\;$ 
converges if $p > 1$, and diverges if $p \le 1$.
E.g. $p = 3$, $a_n = \dfrac{1}{n^3}\implies \sum_{n=1}^\infty \dfrac{1}{n^3}$ converges.
Then $(a_n)^{1/3}$ gives $\sum_{n=1}^\infty \left(\dfrac{1}{n^3}\right)^{1/3} =\;\; \sum_{n=1}^\infty \dfrac{1}{n},\;\;$
which diverges.

Now we need to check whether the fact that $\sum_{n=1}^{\infty}a_n$ converges, implies that $\sum_{n=1}^{\infty}a_n^2$ converges. This is more of a challenge, since it seems to follow that, yes, it must.
Here, we have to get creative to find a counterexample, if one exists: 
If we choose $\sum a_n$ to be an alternating series (sign of terms change depending on even, odd $n$) and so that  $\sum_{n=1}^{\infty}a_n$ would converge but not $\sum_{n=1}^{\infty}a_n^2$, we are in luck. 
Let $ a_n=\dfrac{(-1)^n}{\sqrt{n}}$ (the sum of which converges, non-absolutely, since some terms are positive, others negative). 
Then $(a_n)^2 = \left(\dfrac{(-1)^n}{n^{1/2}}\right)^2 = \dfrac{1}{n}$; this sum, with all positive terms, you will recognize to be the harmonic series, which you know diverges.
A: Youu want to decide whether or not the statements are true for any sequence $(a_n)$. So you either prove them, or provide a counterexample.
What you have done is wrong on many counts. You took a sequence $a_n$ and said (I believe) $\sum_{n=1}^{\infty}a_n$ converges because $a_n\to 0$. This is wrong as the harmonic series $\sum_{n=1}^{\infty}\frac1n$ diverges although $\frac1n\to 0$. 
For an easier counterexample take $a_n=\frac{1}{n^3}$ for the first problem. Indeed,
$\sum_{n=1}^{\infty}\frac1{n^3}$ converges as 
$$0\le \frac1{n^3}\le \frac1{n^2}$$
and the second series converges (why?). You can also use the integral test or the Cauchy Condensation test. What is
$$\sum_{n=1}^{\infty}\left(\frac1{n^3}\right)^{\frac13}?$$
The second problem is trickier: You want $\sum_{n=1}^{\infty}a_n$ to converge but not $\sum_{n=1}^{\infty}a^2_n$. This creates a small problem: 
Because $a_n\to 0$, for large $n$, $\left|a_n\right|\ge a^2_n$ (why?) which should mean that the 
second series would converge. Not so much, if we choose $a_n$ to alternate signs. Then  $\sum_{n=1}^{\infty}a_n$ would converge but not $\sum_{n=1}^{\infty}a^2_n$ as in the second series the terms would only be added, while in the first series 
terms are also subtracted (alternating signs). An example is $a_n=\frac{(-1)^n}{\sqrt{n}}$
A: $\sum\limits_{n=1}^\infty\frac{1}{n^3}$ & $\sum\limits_{n=1}^\infty(-1)^n\frac{1}{\sqrt n}$ are convergent but $\sum\limits_{n=1}^\infty\frac{1}{n}$ is not.
