Infinite series of geometric?

**Anonymous1** · April 12th 2010, 04:37 PM

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

Necessarily,

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

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

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

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

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

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

Thanks.

**234578** · April 12th 2010, 04:51 PM

Originally Posted by Anonymous1

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

Necessarily,

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

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

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

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

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

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

Thanks.

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

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