Question

Show that $1+\frac{1}{3\cdot 2^2}+\frac{1}{5.2^4}+\frac{1}{7.2^6}+....={\log }_{e}3$.

Answer 1

We know from expansions

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....$

$\log (1-x)=-x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....$

So $\log (1+x)+\log (1-x)=2(x+\frac{x^3}{3}+\frac{x^5}{5}+.....)$

For $x=\frac{1}{2}$

$\log \left(\frac{3}{2}\right)-\log \left(\frac{1}{2}\right)=2(\frac{1}{2}+\frac{{\left(\frac{1}{2}\right)}^3}{3}+\frac{{\left(\frac{1}{2}\right)}^5}{5}+.....)$

$\log \left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)=1+\frac{{\left(\frac{1}{2}\right)}^2}{3}+\frac{{\left(\frac{1}{2}\right)}^4}{5}+.....)$

$\log \left(3\right)=1+\frac{1}{3\cdot 2^2}+\frac{1}{5\cdot 2^4}+\frac{1}{7\cdot 2^6}+.....)$