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

  • $\begingroup$ The series is alternating and so is convergent by the Libenitz's test. $\endgroup$ – Albert Dec 23 '15 at 17:35
  • $\begingroup$ @Albert You mean "Leibniz," not "Leibnitz." $\endgroup$ – Mark Viola Dec 23 '15 at 17:45
  • $\begingroup$ @Dr.MV And surely not "Libenitz" ;-) $\endgroup$ – Przemysław Scherwentke Dec 23 '15 at 17:57
  • $\begingroup$ @PrzemysławScherwentke Yes, surely not. ;-)) $\endgroup$ – Mark Viola Dec 23 '15 at 17:59

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.


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