- #1

cbarker1

Gold Member

MHB

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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