Solve $\sum_{i=1}^{\infty}S^i=1$ for $S$ $\sum_{i=1}^{\infty}S^i=1$ for S.  So I'm assuming that they assign S to be 1 and then you just keep increasing the exponent as the term number goes up till infinity. But that would just be 1+1+1... till infinity. What am I doing wrong? 
Thanks! 
 A: In case John is still hopelessly lost, here is a slightly different way of going about it. 
We want to solve $$S+S^2+S^3+\cdots=1$$ Multiply both sides by $S$: $$\qquad S^2+S^3+\cdots=S$$ Subtract the second equation from the first: almost everything on the left cancels out, leaving just $$S=1-S$$ I trust you can take it from there. 
Now I've swept something under the rug here; I've treated infinite sums as if they followed the same rules as finite ones. Sooner or later the difficulty with that will be explained to you, or, if you can't wait, post a new question about it and someone will be happy to answer you here. 
A: Let 
$$T_{k}=S+S^{2}+S^{3}+\ldots +S^{k}\tag{1}$$
be the sum of the first $k$ terms of the geometric progression whose first term is $S$, the ratio is $S$ and the last term is $S^k$. Then 
$$ST_{k}=S^{2}+S^{3}+S^{4}+\ldots +S^{k+1}.\tag{2}$$
Hence $(2)-(1)$ yields:
$$\begin{eqnarray*}
ST_{k}-T_{k} &=&\left( S^{2}+S^{3}+S^{4}+\ldots +S^{k+1}\right) -\left(
S+S^{2}+S^{3}+\ldots +S^{k}\right)  \\
(S-1)T_{k} &=&S^{k+1}-S \\
T_{k} &=&\frac{S^{k+1}-S}{S-1}.
\end{eqnarray*}$$
$$\tag{3}$$
You are given $$\lim_{k\rightarrow \infty }T_{k}=1.\tag{4}$$ If $\left\vert
S\right\vert <1$ then $\lim_{k\rightarrow \infty }S^{k+1}=0$. Thus 
$$\lim_{k\rightarrow \infty }T_{k}=\lim_{k\rightarrow \infty }\frac{S^{k+1}-S}{%
S-1}=\frac{\lim_{k\rightarrow \infty }S^{k+1}-S}{S-1}=-\frac{S}{S-1}=\frac{S%
}{1-S}.\tag{5}$$
Solve 
$$\frac{S}{1-S}=1\tag{6}$$
for $S$, and confirm that $\left\vert S\right\vert <1$.
A: It looks as if you have not quite given the full question.  My guess is that the actual question is something like:
$$\text{Solve}\quad\sum_{i=1}^\infty S^i=1\quad\text{for}\quad S.$$
If that is the case, you want to find all the numbers $S$ such that $\sum_{i=1}^\infty S^i=1$.
So we want to solve the equation
$$S+S^2+S^3+ \cdots = 1.$$
The expression on the left is an infinite geometric series.  From experience you may be able to spot a number  $S$ that "works."  But to make sure you have found all the possible answers (there is actually only one), let's use a formula.
Do you know a simple expression for $\sum_{i=1}^\infty S^i$?  It may help to notice that
$$S+S^2+S^3+\cdots= S(1+S+S^2+\cdots).$$
Do you know a simple expression for $1+S+S^2+\cdots$?
Added:
I will take for granted that you have been given a formula for the sum of an infinite geometric series.  For it seems unlikely you would be asked the question otherwise.  The relevant result is that if $|x|<1$ then
$$1+x+x^2+x^3+\cdots =\frac{1}{1-x}.$$ 
Or maybe the version you were given is that if $|r|<1$ then
$$a+ar+ar^2+ar^3+\cdots=\frac{a}{1-r}.$$
From the first version we get
$$S+S^2+S^3+\cdots=S(1+S+S^2+\cdots)= \frac{S}{1-S}.$$
From the other version, by putting $a=S$ and $r=S$, we get the same result.
So now we want to solve the equation
$$\frac{S}{1-S}=1.$$
We can rewrite this equation as $S=1-S$, and then as $2S=1$. This has the solution $S=1/2$.
Comment: So $1/2$ is the only solution of the equation.  Since $S=1/2$ is a solution, from the original equation we get
$$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots+\frac{1}{2^n}+\cdots=1.$$
This can be interpreted in various ways, the tastiest of which is probably the cake division described in a comment by @Hans Lundmark.
Here is a more boring interpretation.  You start at the origin, and take a step a $1/2$ meter to the right. Now you are at the point with $x$-coordinate $1/2$.  Take a step of $1/4$ m to the right. Now you are at $1/2+1/4$. Take a step of $1/8$ m to the right. Now you are at $1/2+1/4+1/8$.  Continue, forever. After you have taken a large number of steps, say $100$, you will be extremely close to the point with $x$-coordinate $1$. 
A: As Andre points out the notation means we have that S+S^2+...=1.  So, S(1+S+...)=1.  Thus, 1+S+S^2+...=1/S, since it comes as quite clear that S does not equal 0.  So, S+S^2+...=(1/S)-1.  We had that S+S^2+...=1 above, so we can now infer that 1=(1/S)-1.   Thus, 2=(1/S), so S=1/2.  
