Question

Find the value of $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....$

Answer 1

The sum of the series by the binomial theorem, if $0\le x<1,$ we have,

$\frac{1}{(1+x)}=1-x+x^2-x^3+....$

Integrating between limits $0$ to $1$

${\int }_0^1\frac{dx}{(1+x)}={\int }_0^1(1-x{+}^2-x^3+...)$

$\left[\log \right(1+x){]}_0^1={[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+....]}_0^1$

$\Rightarrow$ $\log (1+1)-\log (1-0)=[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..]-0$

$\Rightarrow$ $\log 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....$