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

1. Empty

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

2. jtvatsim

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.

3. anonymous

Ah yes thanks guys!

4. anonymous

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 $$c$$.