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

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

 

[math771]

Hi Cbarker1,

To find M in the definition of convergence for infinite series, you need to use the fact that for any $\varepsilon>0$, there exists an $M\in\Bbb{N}$ such that $k\ge M$, $\left| S_k-0 \right|<\varepsilon$. In this case, we are trying to prove that $\sum_{n=0}^{\infty} r^n=\frac{1}{1-r}$, so we need to show that for any given $\varepsilon>0$, there exists an $M\in\Bbb{N}$ such that $\left| \sum_{n=0}^{k} r^n-\frac{1}{1-r} \right|<\varepsilon$ for all $k\ge M$.

To do this, we can use the proof by induction that you started. You have correctly shown that for $k=1$, $S_1=\sum_{n=0}^{1-1} r^n=\frac{1-r^n}{1-r}$ and $S_1=1=\frac{1-r}{1-r}$. Now, to prove the statement for $k+1$, we need to show that $S_{k+1}=\sum_{n=0}^{k} r^n=\frac{1-r^{1+n}}{1-r}$.

To do this, we can use the fact that $S_k=\sum_{n=0}^{k-1} r^n=\frac{1-r^n}{1-r}$, which you assumed in the proof by induction. Then, we can use the definition of $S_{k+1}$ and substitute in the expression for $S_k$. This will give us:

$S_{k+1}=\sum_{n=0}^{k} r^n=r^k+\sum_{n=0}^{k-1} r^n=r^k+S_k$

Now, we can use the expression for $S_k$ that we assumed in the proof by induction to get:

$S_{k+1}=r^k+\frac{1-r^k}{1-r}=\frac{r^k-r^{k+1}+1-r^k}{1-r}=\frac{1-r^{k+1}}{1-r}$

This is the