# Infinite series of geometric?

• Apr 12th 2010, 04:37 PM
Anonymous1
Infinite series of geometric?
Okay, so I'm trying to find the constant $\displaystyle c,$ but I am rusty on my elementary calculus. Any help is greatly appreciated.

Necessarily,

$\displaystyle [c + c + (\frac{2}{3})^0 c+ (\frac{2}{3})^1c + (\frac{2}{3})^2 c.....] = 1$

$\displaystyle \Rightarrow [2c + \sum_{n=2}^{\infty} (\frac{2}{3})^{n-2}c] = 1$

$\displaystyle \Rightarrow [2 + \sum_{n=2}^{\infty} (\frac{2}{3})^{n-2}] = \frac{1}{c}$

Now, I need to find the sum... Can I reparameterize and let $\displaystyle n-2 = j$ and then find the geo-sum? Like,

$\displaystyle \Rightarrow [2 + \sum_{j=0}^{\infty} (\frac{2}{3})^{j}] = 2+ \frac{1}{1- 2/3} = \frac{1}{c}$

$\displaystyle \Rightarrow c= \frac{1}{5}?$

Thanks.
• Apr 12th 2010, 04:51 PM
234578
Quote:

Originally Posted by Anonymous1
Okay, so I'm trying to find the constant $\displaystyle c,$ but I am rusty on my elementary calculus. Any help is greatly appreciated.

Necessarily,

$\displaystyle [c + c + (\frac{2}{3})^0 c+ (\frac{2}{3})^1c + (\frac{2}{3})^2 c.....] = 1$

$\displaystyle \Rightarrow [2c + \sum_{n=2}^{\infty} (\frac{2}{3})^{n-2}c] = 1$

$\displaystyle \Rightarrow [2 + \sum_{n=2}^{\infty} (\frac{2}{3})^{n-2}] = \frac{1}{c}$

Now, I need to find the sum... Can I reparameterize and let $\displaystyle n-2 = j$ and then find the geo-sum? Like,

$\displaystyle \Rightarrow [2 + \sum_{j=0}^{\infty} (\frac{2}{3})^{j}] = 2+ \frac{1}{1- 2/3} = \frac{1}{c}$

$\displaystyle \Rightarrow c= \frac{1}{5}?$

Thanks.

yep your answer is correct, you can definitely sub in j=n-2 and as n-2 goes to infinity so will j.