2

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 $n$.

ARs
  • 125
  • $$\sum_{1}^n P_0(1+r)^n = {P_0(r^n - 1)\over r -1} $$ will work I guess. –  May 01 '17 at 13:31
  • It is a geometric series, with ratio $1+r$. You want to solve for $n$ such that $\sum_{k=1}^n (1+r)^k= \frac{1}{P_0}$ (be careful, it should be $k$ in the sum, not $n$: the summation index cannot be the same as the upper bound in the sum). Using the formula for geometric sums, this is solving for $n$ such that $$(1+r)\cdot \frac{(1+r)^n-1}{r} = \frac{1}{P_0}$$ – Clement C. May 01 '17 at 13:32

1 Answers1

1

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 $n$.

Marcus Andrews
  • 4,941