find the sum of a series

**twilightstr** · May 14th 2009, 11:25 AM

find the sum of a series sigma from n=0 to infinity of 1/[(2n+1)(2^(n+1))]

**Moo** · May 14th 2009, 11:52 AM

Hello,

Originally Posted by twilightstr

find the sum of a series sigma from n=0 to infinity of 1/[(2n+1)(2^(n+1))]

$\begin{aligned} \sum_{n\geq 0} \frac{1}{(2n+1)2^{n+1}} &=\sum_{n\geq 0}\frac{(1/2)^{n+1}}{2n+1} \\ &=\sum_{n\geq 0} \frac{(1/\sqrt{2})^{2n+2}}{2n+1} \\ &=\frac{1}{\sqrt{2}}\sum_{n\geq 0} \frac{(1/\sqrt{2})^{2n+1}}{2n+1} \\ &=\frac{1}{\sqrt{2}}\sum_{n\geq 0} \int_0^{1/\sqrt{2}} x^{2n} ~dx \end{aligned}$

We can exchange summation and integral (the stuff inside is positive)

$\begin{aligned} \sum_{n\geq 0} \frac{1}{(2n+1)2^{n+1}} &=\frac{1}{\sqrt{2}}\int_0^{1/\sqrt{2}} \left[\sum_{n\geq 0} (x^2)^n\right] ~dx \\ &=\frac{1}{\sqrt{2}}\int_0^{1/\sqrt{2}} \frac{1}{1-x^2} ~ dx \\ &=\frac{1}{\sqrt{2}}\int_0^{1/\sqrt{2}} \frac 12\cdot\left(\frac{1}{1-x}+\frac{1}{1+x}\right) ~dx \\ &=\frac{1}{2\sqrt{2}} \cdot \left. \ln\left(\frac{1+x}{1-x}\right)\right|_0^{1/\sqrt{2}} \end{aligned}$

$S=\boxed{\frac{1}{2\sqrt{2}} \cdot \ln\left(\frac{5+2\sqrt{2}}{3}\right)}$

I hope I didn't make too many mistakes

**Krizalid** · May 14th 2009, 03:41 PM

or $\sum\limits_{n\ge 0}{\frac{1}{(2n+1)2^{n+1}}}=\sum\limits_{n\ge 0}{\frac{2^{n}}{(2n+1)2^{2n+1}}}=\int_{0}^{1/2}{\left( \sum\limits_{n\ge 0}{\left( 2x^{2} \right)^{n}} \right)\,dx}=\int_{0}^{1/2}{\frac{dx}{1-2x^{2}}}.$

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