Infinite Geometric Series

Question

I'm currently stuck on this question:

What is the value of c if $\sum_{n=1}^\infty (1 + c)^{-n}$ = 4 and c > 0?

This appears to be an infinite geometric series with a = 1 and r = $(1 + c)^{-1}$, so if I plug this all into the sum of infinite geometric series formula $S = \frac{a}{1 - r}$, then I get the following:

$4 = \frac{1}{1 - (1 + c)^{-1}}$, which eventually lets me solve c = $\frac{1}{3}$. But this answer isn't right. Can someone help me out? Thanks in advance!

Answer 1

HINT:

As $n$ starts with $1$ not $0$

$\displaystyle a=(1+c)^{-1}=\frac1{1+c},$ not $(1+c)^0=1$

Answer 2

First of all, to converge it must be that

$$\;\frac1{|c+1|}<1\iff |c+1|>1 \iff c>0\;\;or\;\;c<-2\;$$

and applying the well known formula for the sum of a converging gometric sieres we get

$$4=\sum_{n=1}^\infty\frac1{(c+1)^n}=\frac{\frac1{c+1}}{1-\frac1{c+1}}$$

you get the value of $\;c\;$ ...(the first summand is not $\;1\;$ ...)