MHB The proof of the infinite geometric sum

[cbarker1]

Dear Everybody,

I need some help with find M in the definition of the convergence for infinite series.

The question ask, Prove that for $-1<r<1$, we have $\sum_{n=0}^{\infty} r^n=\frac{1}{1-r}$.
Work:
Let $\sum_{n=0}^{k} r^n=S_k$. Let $\varepsilon>0$, we must an $M\in\Bbb{N}$ such that $k\ge M$, $\left| S_k-0 \right|<\varepsilon$. That is $\left|\sum_{n=0}^{k} r^n \right|<\varepsilon$
Finding an M,

$S_k=\sum_{n=0}^{k-1} r^n=\frac{1-r^n}{1-r}$
Proof by Induction
For k=1, $S_1=\sum_{n=0}^{1-1} r^n=\frac{1-r^n}{1-r}$
$S_1=\sum_{n=0}^{0} r^n=1=\frac{1-r}{1-r}$
Assume for all k in the natural numbers, $S_k=\sum_{n=0}^{k-1} r^n=\frac{1-r^n}{1-r}$
Then, Need to show $S_{k+1}=\sum_{n=0}^{k} r^n=\frac{1-r^{1+n}}{1-r}$
Here is where I am stuck.
Thanks,
Cbarker1

 

[Greg]

Re: the proof of the infinite geometric sum

$$\frac{1-r^{n+1}}{1-r}=\frac{(1-r)(1+r+r^2+\cdots+r^{n})}{1-r}$$