# Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series. sum_{n=2}^inftyfrac{(-1)^nn}{n^2-3}

Series
Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
$$\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}$$

2020-10-26

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.