# Integral of geometric series

• Jan 20th 2010, 12:07 AM
ban26ana
Integral of geometric series
http://www.flickr.com/photos/10799158@N07/4289486961/http://farm5.static.flickr.com/4017/...366a8543_m.jpg

Hint: Cal II stuff; what is a geometric series? Try converting the integrand to something simpler.

I don't even know where to begin with this one. The hint isn't helping me.(Wondering) I know what a geometric series is, but I'm not sure how to find the integral of this, or how to convert it into something that I can find the integral.
• Jan 20th 2010, 12:17 AM
Jhevon
Quote:

Originally Posted by ban26ana
http://www.flickr.com/photos/10799158@N07/4289486961/http://farm5.static.flickr.com/4017/...366a8543_m.jpg

Hint: Cal II stuff; what is a geometric series? Try converting the integrand to something simpler.

I don't even know where to begin with this one. The hint isn't helping me.(Wondering) I know what a geometric series is, but I'm not sure how to find the integral of this, or how to convert it into something that I can find the integral.

You know what is a geometric series, so I suppose you know the formula to compute it. use that formula to convert the series into an algebraic expression which you can integrate. the series in the integral is actually equal to $\frac 1{1 - x}$
• Jan 20th 2010, 12:18 AM
Prove It
Quote:

Originally Posted by ban26ana
http://www.flickr.com/photos/10799158@N07/4289486961/http://farm5.static.flickr.com/4017/...366a8543_m.jpg

Hint: Cal II stuff; what is a geometric series? Try converting the integrand to something simpler.

I don't even know where to begin with this one. The hint isn't helping me.(Wondering) I know what a geometric series is, but I'm not sure how to find the integral of this, or how to convert it into something that I can find the integral.

A geometric series $\sum_{k = 0}^{\infty}ar^{k} = \frac{a}{1 - r}$ as long as $|r| < 1$.

So in your case $\sum_{k = 0}^\infty x^k = \frac{1}{1 - x}$, provided $|x| < 1$.

From the limits of your integral, we can see that $0 \leq x \leq \frac{1}{2}$, so the condition $|x| < 1$ is satisfied.

So you are really trying to evaluate

$\int_0^{\frac{1}{2}}{\frac{1}{1 - x}\,dx}$
• Jan 20th 2010, 12:49 AM
ban26ana
Thank you for showing me the obvious.(Headbang) For some reason I was confusing x with k, and trying to put the values 0 and 1/2 as k values...we'll just say it's almost 4 am and I may need some coffee. Thanks!