# Geometrical series extra term confusion

Given $$P = \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-1}} + \frac{A}{(1+x)^{N}},$$

how can I arrive at the textbook's expression $$P = A\left( \frac{1 - \frac{1}{(1+x)^N}}{x}\right)$$

### Attempt at solution

This seems like a divergent geometrical series, I know that the partial sum $$S_n = \frac{1-r^N}{1 - r},$$ where $$r$$ is the common ratio $$1 / (1+x)$$, but when I write down $$S_n$$, I find an extra $$1+x$$ factor $$P = A\frac{S_n}{1+x}.$$ It led me to think that maybe because in the definition of $$S_n$$ for an infinite geometrical series we start at $$x^0$$, i.e., $$a + ax + ax^2 + \dots$$, and in this problem we start at $$ax + ax^2 + \dots$$, then there is some term missing around, but I can't seem to make it work.

Can anyone point out what I am missing ?

• $a+ar+ar^2+\cdots +ar^{n-1} = a\frac{1-r^n}{1-r}$ so $ar+ar^2+\cdots +ar^{n-1}+ar^{n} = ar \frac{1-r^n}{1-r}$ Jun 24 at 15:23
• You may also pick common ratio $=(1+x)$ and the first term $=\frac{A}{(1+x)^N}$, then the denominator $x$ in the textbook answer becomes obvious. Jun 24 at 15:34

Here, we have $$r=\frac{1}{x+1}$$ is the common ratio. Next, we have \begin{align} P&:=Ar+Ar^2+\cdots +Ar^N\\ &=Ar(1+r+\cdots +r^{N-1})\\ &=Ar\left(\frac{1-r^{(N-1)+1}}{1-r}\right)\\ &=Ar\frac{1-r^N}{1-r}. \end{align} Plugging in $$r=\frac{1}{x+1}$$ gives the desired answer. The simple trick here to deal with the fact that the leading term is $$Ar$$ rather than 1, is to just factor out that 'troublesome' term.
$$P = \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-1}} + \frac{A}{(1+x)^{N}}$$
$$=\frac{A}{1+x}(1 + \frac{A}{1+x} + \frac{A}{(1+x)^2} + \dots + \frac{A}{(1+x)^{N-2}} + \frac{A}{(1+x)^{N-1}})$$
$$=\frac{A}{1+x}(\frac{1-\frac{1}{(1+x)^N}}{1-\frac{1}{1+x}})$$
$$= \frac{A}{1+x}(\frac{1-\frac{1}{(1+x)^N}}{\frac{x}{1+x}})$$
$$=A(\frac{1-\frac{1}{(1+x)^N}}{x})$$