Answer to Question #103849 in Calculus for BIVEK SAH

Question

Test the absolute and conditional convergence of the series

Infinity
∑ (-1)^n/(2n+1)
n=1

Answer 1

$\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{2n+1}, (1)\\ a_n=\frac{(-1)^n}{2n+1}, |a|_n=\frac{1}{2n+1}.$

Consider a series  of modules

$\sum\limits_{n=1}^{\infty}|a_n|=\sum\limits_{n=1}^{\infty}\frac{1}{2n+1}, (2)\\$

consider a harmonic series

$\sum\limits_{n=1}^{\infty}\frac{1}{n}, b_n=\frac{1}{n}.\\$

It is divergent. Consider the limit

$\lim\limits_{n\mapsto \infty}\frac{|a_n|}{b_n}=\lim\limits_{n\mapsto \infty}\frac{n}{2n+2}=\frac{1}{2},$

hence the series of modules (2) is divergent.

We investigate the series (1) on the basis of Leibniz criterion

$1) |a_1|=\frac{1}{3}>|a_2|=\frac{1}{5}>|a_3|=\frac{1}{7}>...>\\ >|a_n|=\frac{1}{2n+1}>...\\ 2) \lim\limits_{n\mapsto \infty}|a_n|=\lim\limits_{n\mapsto \infty}\frac{1}{2n+1}=0$

Therefore, the series (1) is convergent on the basis of Leibniz criterion.

Since the series of modules (2) is divergent and the series (1) is convergent according to Leibniz criterion, the series (1) is conditionally convergent