Question

$\sum _{n=1}^{\infty }\frac{\left(-1\right)^n}{n}$

Check it

Answer 1

$\mathrm{Check\:convergence\:of\:}\sum _{n=1}^{\infty \:}\frac{\left(-1\right)^n}{n}:\quad \mathrm{converges}$

$Steps$

$\sum _{n=1}^{\infty \:}\frac{\left(-1\right)^n}{n}$

$\mathrm{Apply\:Alternating\:Series\:Test}:\quad \mathrm{converges}$

$\mathrm{Suppose\:that\:for\:}a_n\mathrm{,\:there\:exists\:an\:}N\mathrm{\:so\:that\:for\:all\:}n\ge N$

$a_n\mathrm{\:is\:positive\:and\:monotone\:decreasing\:}$

$\lim _{n\to \infty }a_n=0$

$\mathrm{Then\:the\:alternating\:series\:}\sum \left(-1\right)^na_n\mathrm{\:and\:}\sum \left(-1\right)^{n-1}a_n\mathrm{\:both\:converge}$

$\mathrm{By\:the\:alternating\:series\:test\:criteria}$

$=\mathrm{converges}$

Answer 2

Explanation:

$\mathrm{Check\:convergence\:of\:}\sum _{n=1}^{\infty \:}\frac{\left(-1\right)^n}{n}:\quad \mathrm{converges}$

$Steps$

$\sum _{n=1}^{\infty \:}\frac{\left(-1\right)^n}{n}$

$\mathrm{Apply\:Alternating\:Series\:Test}:\quad \mathrm{converges}$

$\mathrm{\sf Then\:the\:alternating\:series\:}\sum \left(-1\right)^na_n\mathrm{\:and\:}\sum \left(-1\right)^{n-1}a_n\mathrm{\:both\:converge}$

$=\mathrm{converges}$