# The proof of the infinite geometric sum

• MHB
• cbarker1
In summary: S_{k+1}=\sum_{n=0}^{k} r^n=\frac{1-r^{k+1}}{1-r}$. This completes the proof by induction.In summary, to prove the convergence of infinite series, we need to find an$M\in\Bbb{N}$such that for any$\varepsilon>0$,$\left| \sum_{n=0}^{k} r^n-\frac{1}{1-r} \right|<\varepsilon$for all$k\ge M$. This can be done using the proof by induction, as shown above cbarker1 Gold Member MHB Dear Everybody, I need some help with find M in the definition of the convergence for infinite series. The question ask, Prove that for$-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 Re: the proof of the infinite geometric sum $$\frac{1-r^{n+1}}{1-r}=\frac{(1-r)(1+r+r^2+\cdots+r^{n})}{1-r}$$ 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}\$

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## 1. What is the formula for finding the sum of an infinite geometric series?

The formula for finding the sum of an infinite geometric series is S = a / (1 - r), where a is the first term and r is the common ratio.

## 2. How do you know if an infinite geometric series has a finite sum?

An infinite geometric series has a finite sum if the absolute value of the common ratio (r) is less than 1. If the absolute value of r is greater than or equal to 1, the series will not have a finite sum.

## 3. Can the sum of an infinite geometric series be negative?

Yes, the sum of an infinite geometric series can be negative if the common ratio (r) is negative and the first term (a) is also negative. However, if both a and r are positive, the sum will always be positive.

## 4. Can you use the formula for finding the sum of an infinite geometric series if the common ratio is 1?

No, the formula for finding the sum of an infinite geometric series only works when the absolute value of the common ratio (r) is less than 1. If r = 1, the series will not have a finite sum.

## 5. How is the formula for finding the sum of an infinite geometric series derived?

The formula for finding the sum of an infinite geometric series is derived using the concept of limits in calculus. It involves taking the limit of the partial sums of the series as the number of terms approaches infinity.

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