# Solving an infinite series that doesn't seem to be harmonic or geometric

• Nov 7th 2010, 04:28 PM
funnyguy
Solving an infinite series that doesn't seem to be harmonic or geometric
I'm solving a visual problem, and sparing the boring details, I've found the sequence to be:

${a_n} = \frac{1}{2^{2n-1}}$

I need to solve for the sum of the series from 1 to infinity. Wolfram easily solves this as 2/3. However, I've poured over 2 calculus books and I've enlisted help but we can't solve it by hand.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}$

Any help at simplifying this into a series which I can solve would be greatly appreciated.
• Nov 7th 2010, 05:06 PM
skeeter
Quote:

Originally Posted by funnyguy
I'm solving a visual problem, and sparing the boring details, I've found the sequence to be:

${a_n} = \frac{1}{2^{2n-1}}$

I need to solve for the sum of the series from 1 to infinity. Wolfram easily solves this as 2/3. However, I've poured over 2 calculus books and I've enlisted help but we can't solve it by hand.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{2n-1}}$

Any help at simplifying this into a series which I can solve would be greatly appreciated.

$\displaystyle \frac{1}{2^{2n-1}} = 2^{1-2n} = \frac{2}{2^{2n}} = \frac{2}{4^n}$

$\displaystyle 2 \sum_{n=1}^{\infty} \frac{1}{4^n} = 2 \cdot \frac{\frac{1}{4}}{1 - \frac{1}{4}} = 2 \cdot \frac{1}{3} = \frac{2}{3}$