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

My attempt: 

Let $c = n^q$ 

Then since, $$\sum_{i=m}^{n} (ca_i) = c(\sum_{i=m}^{n} a_i),$$ we can write

$$c\sum_{n=1}^{\infty} x^{n}$$
and by a Lemma concerning geometric series, if $|x|&lt;1$ then $$\sum_{n=1}^{\infty} x^{n}$$ is  absolutely convergent. If it is absolutely convergent it is also conditionally convergent. Now I can use the zero test and we see that the limit of $x^n$ as n->infinity = 0.
Our series has $n$-th term $a_n=n^qx^n$ and thus$$\frac{a_{n+1}}{a_n}=x\cdot \left(\frac{n+1}{n}\right)^q=x\cdot \left( 1+\frac1n \right)^q$$

It follows that $\lim\limits_{n\to\infty} \left \lvert \frac{a_{n+1}}{a_n}\right \rvert=\lvert x \rvert &lt;1$, so by the ratio test the series $\sum\limits_{n\geq 1} a_n$ converges absolutely.

EDIT: Granted, if $\sum\limits_{n\geq 1} \lvert a_n \rvert$ converges, then $\lvert a_n\rvert \to 0$, so $\lim\limits_{n \to \infty} n^qx^n=0$ whenever $\lvert x \rvert &lt;1$.
By ration test we examine

$$
\lim_{n\rightarrow +\infty} \frac{a_n}{a_{n+1}}=\lim_{n\rightarrow +\infty} \frac{n^q}{(n+1)^q}=\lim_{n\rightarrow +\infty} \left(\frac{n}{n+1}\right)^q=1
$$

Here we actually applied Cauchy-Haramard theorem

So we know that this power series absolutely converges on (-1,1) (on -1,1 the series obviously diverges)