We have given series

\(\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}\)

Let, \(\sum_{n=2}^\infty(-1)^na_n=\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}\)

Then, \(a_n=\frac{n}{n^2-3}\)

\(a_{n+1}=\frac{n+1}{(n+1)^2-3}\)

\(\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{n}{n^2-3}\)

\(=\lim_{n\rightarrow\infty}\frac{n}{n^2(1-\frac{3}{n^2})}\)

\(=0\)

\(n^2-3<(n+1)^2-3\)

\(\frac{1}{n^2-3}>\frac{1}{(n+1)^2-3}\)

\(\frac{n}{n^2-3}>\frac{n}{(n+1)^2-3}\)

\(a_n>a_{n+1}\)

Therefore, sequence \(\left\{a_n\right\}\) is decreaing.

So, by alternating series test given series is convergent.