For what real values of $x$ does the series $\sum_{n=1}^{\infty} \frac{n}{5^{n-1}}(x+2)^n $ converge? 
For what real values of $x$ does the series $\sum_{n=1}^{\infty} \frac{n}{5^{n-1}}(x+2)^n$ converge?

My attempt: $-1< (x+2)^n < 1$. Now, $-3 < x < -1$, so by the Leibniz test,   the given series will converge in $-3 <x<-1$.
Is this correct?

HINT
By ratio test
$$\left| \frac{(n+1)(x+2)^{n+1}}{5^{n}}\frac{5^{n-1}}{n(x+2)^n}\right|=\frac{n+1}{5n}\left|x+2\right|\to\frac{1}{5}\left|x+2\right|$$
then the series converges for $|x+2|<5 \implies -7<x<3$.
Then check separetely the cases $x=-7$ and $x=3$.
