Answer
$$14/5\leq x\leq 16/5$$
Work Step by Step
Given
$$ \sum_{n=1}^{\infty} \frac{(-5)^{n}(x-3)^{n}}{n^{2}}$$
Since $a_n = \dfrac{(-5)^{n}(x-3)^{n}}{n^{2}} $ and $a_{n+1} = \dfrac{(-5)^{n+1}(x-3)^{n+1}}{(n+1)^{2}}$, then
\begin{aligned} \rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\ &=\lim _{n \rightarrow \infty}\left|\frac{-5(-5)^{n} \frac{(x-3)(x-3)^{n}}{(n+1)^{2}}}{(-5)^{n} \frac{(x-3)^{n}}{n^{2}}}\right|\\ &=\lim _{n \rightarrow \infty}\left|-5(x-3) \cdot \frac{n^{2}}{(n+1)^{2}}\right|\\ & =5|(x-3)| \lim _{n \rightarrow \infty} \frac{1}{\left(1+n^{-2}\right)^{2}}\\ &=5|(x-3)| \end{aligned}
Then the series converges for $$ 5|(x-3)|<1 \ \to \ 14/5 \lt x \lt 16/5 $$
Test the end points. For $x= 16/5 $
$$ \sum_{n=1}^{\infty} \frac{(-5)^{n}(x-3)^{n}}{n^{2}}= \sum_{n=1}^{\infty}(-1)^{n} \frac{1}{n^{2}}$$ This is alternating series, such that $a_{n}$ decreases and $\lim _{n \rightarrow \infty}\left|a_{n}\right|=0$, which converges.
For $x= 14/5 $
$$ \sum_{n=1}^{\infty} \frac{(-5)^{n}(x-3)^{n}}{n^{2}}= \sum_{n=1}^{\infty}\frac{1}{n^{2}}$$ This is a $p-$series, ($p=2$) , which converges.
Hence, the interval of convergence is $$14/5\leq x\leq 16/5$$
