I am revising for an exam and I am facing the following statement: $\frac{1}{1-p}=\sum_{k=0}^\infty {p^k}$
Unfortunately, I was not able to figure out why this is the case. May somebody of you quickly help me? I appreciate it a lot for a detailed explanation!
From the equaility $(1-p)(1+p+p^2+ \dots + p^k)=1-p^{k+1}$ we deduce $\frac{1-p^{k+1}}{1-p}=1+p+\dots+p^k$
So if $p<1$ we have $\frac{1}{1-p}=\lim\limits_{k\to \infty} \frac{1-p^{k+1}}{1-p}=\lim\limits_{k\to \infty}1+p+p^2+\dots+ p^k$
Since$$\frac{1-p^n}{1-p}=1+p+p^2+...+p^{n-1}=\sum_{i=1}^{n-1}p^k$$
If $|p|<1$, $n\to \infty$, then $$\lim_{n\to \infty}\frac{1-p^n}{1-p}=\frac{1}{1-p}=\sum_{i=1}^{\infty}p^k$$
HINT: consider the finite sum $$\sum_{k=0}^np^k={\frac {{p}^{n+1}}{p-1}}- \left( p-1 \right) ^{-1}$$ and compute the limit for $n$ tends to infinity.
