Suppose $\sum a_n $ converges . Show $\sum n^k a_n$ converges. Suppose that the series of positive numbers $\sum a_n$ can be shown to converge using either the Ratio Test or the Root Test. Show that the series $\sum n^k a_n$ converges for any positive integer $k$. 
Is my proof sufficient?
Proof:
Assume $\sum a_n$ converges by the Ratio Test.
That is $R<1$ for $\varlimsup_{n\to\infty} \frac{a_{n+1}}{a_n}$.
Consider $\sum n^k a_n$, $k\in \mathbb{Z^+}$.
We have $\varlimsup_{n\to\infty} \frac{(n+1)^k a_{n+1}}{n^k a_n}$.
By Induction, $\sum \frac{(n+1)^k a_{n+1}}{n^k a_n} \leq$ $\max\{(n+1)^k\}$ $\frac{a_{n+1}}{a_n} < 1$ , for sufficiently large $k$.
Hence by Ratio Test, $\sum n^k a_n$ converges. 
 A: As it stands, this is not sufficient. For example, you have the line

That is $R<1$ for $\varlimsup_{n\to\infty} \frac{a_{n+1}}{a_n}$.

What is $R$, and what does it have to do with the terms? I assume that you want to define $R$ to be this limit supremum.

We have $\varlimsup_{n\to\infty} \frac{(n+1)^k a_{n+1}}{n^k a_n}$.

I assume that you're aiming to have a bound on this quantity, but you don't. The bound here ought to be $R$, and that needs to be proven - you'll have to study the quantity $$\frac{(n + 1)^k}{n^k}$$ carefully to conclude this.
Then you have

By Induction, $\sum \frac{(n+1)^k a_{n+1}}{n^k a_n} \leq$ $\max\{(n+1)^k\}$ $\frac{a_{n+1}}{a_n} < 1$ , for sufficiently large $k$.

This doesn't really make any sense. If $k$ gets large, the quantity in the maximum gets larger, not smaller, and your bound doesn't work. Moreover, you're using $n$ as the variable of summation and then as something fixed, so I'm really not even sure how to parse this. It's also not clear what induction has to do with any of this.

As a very brief sketch of how to fix this: If you can show that
$$\sum_{n = 1}^{\infty} n^k R^n$$
converges for any $|R| < 1$, you're pretty much done - then some sort of comparison test will finish it. This can be done with the ratio test.
A: Hint:
T. Bongers makes good points. Here is a hint to salvage your proof.
$$\lim_{n\to\infty}\frac{(n+1)^k |a_{n+1}|}{n^k |a_k|}=\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^k\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$
A: First note that (as @T.Bongers pointed out) if $|R|<1$ and $k$ is a positive integer, then $$\lim_{n\to\infty}  |n^kR^n|^{\frac1n}=R\lim_{n\to\infty}n^{\frac kn}=R\left(\lim_{n\to\infty}n^{\frac1n} \right)^k=R<1,$$ so by the root test, $\sum_n n^k R^n$ converges absolutely.
Now, suppose $\{a_n\}$ is a sequence of real numbers such that $$\limsup_{n\to\infty} |a_n|^{\frac1n}=\rho\in(0,1). $$ Let $R\in(\rho,1)$. Choose $N$ so that $n\geqslant N$ implies $|a_n|^{\frac1n}<R$, and hence $|a_n|<R^n$. It follows that $n^k|a_n|\leqslant n^kR^n$ for $n\geqslant N$, and so by basic comparison, $\sum_n n^k a_n$ converges absolutely.
Since $$\limsup_{n\to\infty} c_n^{\frac1n}\leqslant\limsup_{n\to\infty}\frac{c_{n+1}}{c_n} $$ for any positive sequence $\{c_n\}$, whenever the ratio test shows convergence, the root test does too, and so we need only consider the root test here.
