Show that $\sum_{n = 1}^\infty n^qx^n$ is absolutely convergent, and that $\lim_{n \rightarrow \infty}$ $n^qx^n = 0$

I'm having trouble with proving the following for my math study:

Let $x$ be a real number with $|x| < 1$, and $q$ be a real number. Show that the series $\sum_{n = 1}^\infty n^qx^n$ is absolutely convergent, and that $\lim_{n \rightarrow \infty}$ $n^qx^n = 0$

I tried this:

$\sum_{n = 1}^\infty n^qx^n$ = $x + 2^q + 3^qx^3 + ...$
$= x(1 + 2^qx + 3^qx^2 + ...)$

So far, I don't know how I have to do this, I tried to apply the root test on this, but for that I would need to compute the lim sup of the series, so I thought that was not correct. Could you explain me how to prove this?

Thanks in advance!

Solution 1:

See in the root test you need to compute $\limsup_\limits{n\to \infty}|n^qx^n|^{\frac{1}{n}}$.

Now $\limsup_\limits{n\to \infty}|n^qx^n|^{\frac{1}{n}} = \limsup_\limits{n\to \infty}|n^q|^{\frac{1}{n}}|x| = \left(\limsup_\limits{n\to \infty}n^{\frac{1}{n}}\right)^q|x|=|x|<1$ because $\limsup_\limits{n\to \infty}n^{\frac{1}{n}} = 1.$ The result now follows from the Root Test.