Question

The sum of the infinite series

$[1+\frac{\sqrt{2}-1}{2\sqrt{2}}+\frac{3-2\sqrt{2}}{12}+\frac{5\sqrt{2}-7}{24\sqrt{2}}+\frac{17-2\sqrt{2}}{80}+...\infty ]$ is $a-\frac{1}{b}(\sqrt{c}+1){\log }_{e}2$

Find $(a+b{)}^2+c^2$

Answer 1

Let $\frac{\sqrt{2}-1}{\sqrt{2}}=x$

Then $x^2=\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}{)}^2}$.

$=\left(\frac{3-2\sqrt{2}}{2}\right)$;

$x^3={\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}^3=\frac{5\sqrt{2}-7}{2\sqrt{2}}$;

$x^4={\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}^4=\frac{17-12\sqrt{2}}{4}$ and so on.

$\therefore$ Given series

$=1+\frac{1}{2}x+\frac{1}{6}{x}^2+\frac{1}{12}{x}^3+\frac{1}{20}{x}^4+....$

$=1+(1-\frac{1}{2})x+(\frac{1}{2}-\frac{1}{3})x^2+(\frac{1}{3}-\frac{1}{4})x^3+(\frac{1}{4}-\frac{1}{5})x^4+...$

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

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

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

$=\frac{1}{x}[2x+{\log }_e(1-x\left)\right]-{\log }_e(1-x)$

$=2+\frac{1}{x}{\log }_e(1-x)-{\log }_e(1-x)$

$=2+(\frac{1}{x}-1){\log }_e(1-x)$

Putting $x=\frac{\sqrt{2}-1}{\sqrt{2}}$

$=2+(\frac{\sqrt{2}}{\sqrt{2}-1}-1){\log }_e(1-\frac{\sqrt{2}-1}{\sqrt{2}})$

$=2+\left(\frac{1}{\sqrt{2}-1}\right){\log }_e\left(\frac{1}{\sqrt{2}}\right)$

$=2+(\sqrt{2}+1){\log }_e{2}^{-1/2}$

$=2-\frac{1}{2}(\sqrt{2}+1){\log }_{e}2$

$\Rightarrow (a+b{)}^2+c^2=20$