Question

Prove that
$1-\frac{1}{2}+\frac{1}{2}+\frac{1.3}{2.4}.\frac{1}{2^2}-\frac{\mathrm{1.3.5}}{\mathrm{2.4.6}}.\frac{1}{2^3}+\frac{\mathrm{1.3.5.7}}{\mathrm{2.4.6.8}}.\frac{1}{2^4}-...=\sqrt{\frac{2}{3}}$

Answer 1

$y=1-\frac{1}{2.2}+\frac{1.3}{{\mathrm{2.4.2}}^2}-\frac{\mathrm{1.3.5}}{{\mathrm{2.4.6.2}}^3}+.......$

above series can be manipulated as

$y=1-\frac{1}{2.2}+\frac{1.3}{\mathrm{2.2.2}!{.2}^2}-\frac{\mathrm{1.3.5}}{\mathrm{2.2.2.3}!{.2}^3}+.......$

this cleary of the form

$(1+x{)}^n=1+nx+(n\left)\right(n-1)\frac{x^2}{2!}+......(n!)\frac{x^n}{n!}$

Where n is negative or fraction

where $x=\frac{1}{2}$ & $n=-\frac{1}{2}$

$y=(1+\frac{1}{2}{)}^{\frac{-1}{2}}$

$=(\frac{3}{2}{)}^{\frac{-1}{2}}=\sqrt{\frac{2}{3}}$