Solve geometric series? I think this is a geometric series, Im in way over my head here and not even sure of the vocab or how to ask really.
I have
$$\sum_{1}^n P_0(1+r)^n$$
And when $r=.1$ and $P_0=1$ I would like to know how to find $n$ when the $\sum_{1}^n P_0(1+r)^n=10$
I hope I formulated the question correctly, its been over ten years since doing any math work like this. The question comes from my desire to find out when I can expect to get my money back, sum of earnings per share, if the share starts with 1 dollar of earnings, grows earnings at 10%, and I paid $10 for the share. I was able to come up with the function $P_0(1+r)^n$ which will correctly tell me what the earnings will be given some n. I look at those earnings as money returned and so I want to sum them up and know when the sum equals what I paid for the share. But, I have no idea where to start to solve the sum for $n$.

I believe you are looking for $S = \sum_{k=1}^{n} P_0(1+r)^k$, as opposed to $S = \sum_{k=1}^{n} P_0(1+r)^n$
Then $S = P_0(1+r) + P_0(1+r)^2 + P_0(1+r)^3 + ... + P_0(1+r)^n$
Let $(1+r)S = P_0(1+r)^2 + P_0(1+r)^3 + P_0(1+r)^4 + ... + P_0(1+r)^{n+1}$
So $S - (1+r)S = P_0(1+r) - P_0(1+r)^{n+1}$
Rearranging and solving for $S$, we get $S = \dfrac{P_0(1+r)((1+r)^n - 1)}{r}$
Set equal to $10$ and solve for $n$.
