Answer to Question #103940 in Calculus for BIVEK SAH

Question

Test the conditional convergence of the series


1-1/root2+1/root3-1/root4+....

Answer 1

$1-\frac{1}{\sqrt2}+\frac{1}{\sqrt3}-\frac{1}{\sqrt4}+...+\frac{(-1)^{n-1}}{\sqrt n}+...(*)\\ a_n=\frac{(-1)^{n-1}}{\sqrt n}=\frac{(-1)^{n-1}}{ n^\frac{1}{2}}.$

Consider a series of modules

$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+...+\frac{1}{\sqrt n}+...(**)\\$

Series

$\sum\limits_{n=1}^{\infty} \frac{1}{n^\alpha}$

convergent if $\alpha>1$ , divergent if $\alpha \leq1$ .

In our case $\alpha=\frac{1}{2}<1$ then series (**) is divergent.

We examine the series (*) on the basis of Leibniz

$1) |a_1|>|a_2|>...>|a_n|>...$

in our case

$1>\frac{1}{\sqrt 2}>\frac{1}{\sqrt 3}>...>\frac{1}{\sqrt n}>...\\ 2) \lim\limits_{n\mapsto \infty}|a_n|=\lim\limits_{n\mapsto \infty} \frac{1}{\sqrt n}=0.$

According to Leibniz, the series (*) is convergent.

 Since the series (**) is divergent and the series (*) is convergent, the series (*) is conventionally convergent.