Answer
Radius of convergence: $5$
Interval of convergence: $(-5, 5)$
Work Step by Step
If $a_n = \frac{n}{5^n} x^n$, then $\lim\limits_{n \to \infty} |\frac{a_n}{a_{n+1}}| = \lim\limits_{n \to \infty} |\frac{(n+1)x^{n+1}}{5^{n+1}}*\frac{5^n}{nx_n}|$
$ = \lim\limits_{n \to \infty} |\frac{(n+1)x}{5n}| = \lim\limits_{n \to \infty} (\frac{1}{5} + \frac{1}{5n})|x| = \frac{|x|}{5}$.
By Ratio Test, series $\Sigma_{n=1}^{\infty} \frac{n}{5^n} x^n$ converges when $\frac{|x|}{5}\lt1 \Leftrightarrow |x| \lt 5$ so radius of convergence is 5.
When $x = \pm 5$, both series $\Sigma_{n=1}^{\infty} \frac{n(\pm 5)^{2}}{5^n} = \Sigma_{n=1}^{\infty} (\pm 1)^{n}n$ diverge by Test of Divergence, since $\lim\limits_{n \to \infty} |(\pm 1)^{n}n| = \infty$.
Thus, the interval of convergence is $(-5, 5)$.
