find the sum of a series sigma from n=0 to infinity of 1/[(2n+1)(2^(n+1))]
Hello,
$\displaystyle \begin{aligned}
\sum_{n\geq 0} \frac{1}{(2n+1)2^{n+1}}
&=\sum_{n\geq 0}\frac{(1/2)^{n+1}}{2n+1} \\
&=\sum_{n\geq 0} \frac{(1/\sqrt{2})^{2n+2}}{2n+1} \\
&=\frac{1}{\sqrt{2}}\sum_{n\geq 0} \frac{(1/\sqrt{2})^{2n+1}}{2n+1} \\
&=\frac{1}{\sqrt{2}}\sum_{n\geq 0} \int_0^{1/\sqrt{2}} x^{2n} ~dx \end{aligned}$
We can exchange summation and integral (the stuff inside is positive)
$\displaystyle \begin{aligned}
\sum_{n\geq 0} \frac{1}{(2n+1)2^{n+1}}
&=\frac{1}{\sqrt{2}}\int_0^{1/\sqrt{2}} \left[\sum_{n\geq 0} (x^2)^n\right] ~dx \\
&=\frac{1}{\sqrt{2}}\int_0^{1/\sqrt{2}} \frac{1}{1-x^2} ~ dx \\
&=\frac{1}{\sqrt{2}}\int_0^{1/\sqrt{2}} \frac 12\cdot\left(\frac{1}{1-x}+\frac{1}{1+x}\right) ~dx \\
&=\frac{1}{2\sqrt{2}} \cdot \left. \ln\left(\frac{1+x}{1-x}\right)\right|_0^{1/\sqrt{2}} \end{aligned}$
$\displaystyle S=\boxed{\frac{1}{2\sqrt{2}} \cdot \ln\left(\frac{5+2\sqrt{2}}{3}\right)}$
I hope I didn't make too many mistakes