Convergence of the series Find the convergence and absolute convergence of the series $∑\frac{(-1)^{n+1} n}{1+n^2}$

For Absolute convergence, I found out by comparison tests, it is not absolutely convergent. But I couldnt find for convergence. Please help me
Let $a_n=\frac{(-1)^n n}{1+n^2}$. Then $a_{2n}\geq 0$ and $a_{2n-1}\leq 0$ for all $n$.

$|a_n|=\frac{ n}{1+n^2}$. Need to show $|a_n|\geq |a_{n+1}|$ for all $n$. i. e. need $\frac{ n}{1+n^2}\geq \frac{ n+1}{1+(1+n)^2}$. Here, backward method will work. So starting from $n^2+n\geq 1$, you can show it.

Now, $\lim_{n\rightarrow \infty} a_n=0$. So by alternating series test(Leibnitz test) the series converges.
Let $a_n=\frac{(-1)^n n}{1+n^2}$. Then $a_{2n}\geq 0$ and $a_{2n-1}\leq 0$ for all $n$.

$|a_n|=\frac{ n}{1+n^2}$. Need to show $|a_n|\geq |a_{n+1}|$ for all $n$. i. e. need $\frac{ n}{1+n^2}\geq \frac{ n+1}{1+(1+n)^2}$. Here, backward method will work. So starting from $n^2+n\geq 1$, you can show it.

Now, $\lim_{n\rightarrow \infty} a_n=0$. So by alternating series test(Leibnitz test) the series converges.