What do I do, I am unsure how to interpret it as a geometric series. http://i.imgur.com/pyZd2ip.png

Thanks

Question

What do I do, I am unsure how to interpret it as a geometric series. http://i.imgur.com/pyZd2ip.png

Thanks

empty · Answer

You can make a substitution to make it easier perhaps? 

$$x=(1+c)^{-1}$$
$$\sum_{n=1}^\infty x^n =8$$

jtvatsim · Answer

I agree with @empty. You can then use the infinite series formula for geometric series to get a value for x. Then solve for c after. 

That is,
1/(1-x) = 8 solve for x. You will get x = something.
Then,
x = 1/(1+c)
so,     something = 1/(1+c) solve for c.

anonymous · Answer

Ah yes thanks guys!

anonymous · Answer

Let $x = 1+c$
Partial sum is$$s_n = x^{-1} + x^{-2} + x^{=3} + ... + x^{-n}$$Multiply both sides by $x$$$xs_n = 1 + x^{-1} + x^{-2} + ... + x^{-n+1}$$Subtract 1st equation from 2nd equation$$(x-1)s^n = 1 + x^{-n}$$$$s_n = \frac{ 1 + x^{-n} }{ x-1 }$$Sum of infinite series is$$\lim_{n \rightarrow \infty} s_n = \lim_{n \rightarrow \infty} \frac{ 1 + x^{-n} }{ x-1 } = 8$$$$\frac{ 1 }{ x-1 } = 8$$Solve for $x$ and then for $c$.