# What is the general formula for a convergent infinite geometric series?

This question is related, but different, to one of my previous questions (Does this infinite geometric series diverge or converge?). To avoid the previous question getting off-topic, I have created a separate question.

I'm looking for the general formula of a convergent infinite geometric series. I want to be able to calculate any convergent infinite geometric series I come across, regardless of where it starts at. Some examples of this are:

$$\sum_{n=0}^\infty ar^n$$

$$\sum_{n=1}^\infty ar^n$$

$$\sum_{n=2}^\infty ar^n$$

...

$$\sum_{n=5}^\infty ar^n$$

...

etc.

I would appreciate it if someone could present such a formula and explain the reasoning behind it. Also, please illustrate how the formula can be applied to the above examples.

Thank you.

• If you have a formula for the first one, you have it for all of them by simply factoring out the correct power of $r$. And this formula is well known. – Mathematician 42 Nov 7 '16 at 0:29
• @Mathematician42 Can you please elaborate? I have the formula $\dfrac{a-ar^n}{1-r}$. – The Pointer Nov 7 '16 at 0:30
• If $|r|<1$, then $\sum_{n\geq 0} ar^n=a\sum_{n\geq 0}r^n=\frac{a}{1-r}$. Now notice that $\sum_{n\geq m}ar^n=ar^m\sum_{n\geq 0}r^{n}$. – Mathematician 42 Nov 7 '16 at 0:32
• @Mathematician42 I do not understand the second part of your explanation. – The Pointer Nov 7 '16 at 0:35
• We have $\sum_{n\geq m}ar^n=ar^m\sum_{n\geq 0}r^{n}=r^m(a\sum_{n\geq 0}r^n)=r^m\frac{a}{1-r}$. If you understand why $\sum_{n\geq 0}r^n=\frac{1}{1-r}$ when $|r|<1$, then there should be no problem in understanding everything else I'm saying. – Mathematician 42 Nov 7 '16 at 0:39

In general, you have the finite geometric series given by $$\sum\limits_{n=0}^{N-1}ar^n = \frac{a(1-r^N)}{1-r}.$$

Taking the limit of $N\to \infty$ you have the infinite geometric series given by $$\sum\limits_{n=0}^\infty ar^n = \frac{a}{1-r}$$ which converges if and only if $|r|<1$. Now we will consider starting index $N$ instead, i.e. $\sum\limits_{n=N}^\infty ar^n$.

Notice that

$$\sum\limits_{n=0}^\infty ar^n = \sum\limits_{n=0}^{N-1} ar^n + \sum\limits_{n=N}^\infty ar^n = \frac{a}{1-r}$$ and by isolating the desired term we get $$\sum\limits_{n=N}^\infty ar^n = \frac{a}{1-r} - \sum\limits_{n=0}^{N-1} ar^n.$$ The last term is exactly the finite geometric series and hence we get $$\sum\limits_{n=N}^\infty ar^n = \frac{a}{1-r} - \frac{a(1-r^N)}{1-r}.$$

Simplifying we get $$\bbox[5px,border:2px solid red]{\sum\limits_{n=N}^\infty ar^n = \frac{ar^N}{1-r}.}$$

• Exactly correct. That being said, I find it much much easier to remember as $\frac{\text{first term}}{1-\text{common ratio}}$ – erfink Nov 7 '16 at 1:06
• Great explanation. Thank you very much. – The Pointer Nov 7 '16 at 1:13

$$\sum_{n=N}^\infty ar^n=\sum_{n=0}^\infty ar^{n+N}=r^N\sum_{n=0}^\infty ar^n=r^N\left(\frac{a}{1-r}\right)=\frac{ar^N}{1-r}\ \forall\ |r|<1$$

The first step was re-indexing: $\displaystyle\sum_{n=N}^\infty b_n=b_N+b_{N+1}+\dots=\sum_{n=0}^\infty b_{n+N}$

The second step was factoring the $r^N$ term out.

The last steps was using well-known geometric series formula, followed by some algebra and checking convergence.

In my opinion, the simplest way to memorize the formula is

$$\frac{\text{first}}{1 - \text{ratio}}$$

So whether you're computing

$$\frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \ldots$$

or

$$\sum_{n=3}^{\infty} 2^{-n}$$

or

$$\sum_{n=0}^{\infty} \frac{1}{8} 2^{-n}$$

you can quickly identify the sum as

$$\frac{ \frac{1}{8} }{1 - \frac{1}{2}}$$

Similarly, for a finite geometric sequence, a formula is

$$\frac{\text{first} - \text{(next after last)}}{1 - \text{ratio}}$$

The infinite version can be viewed as a special case, where $(\text{next after last}) = 0$.

I find this formula more convenient written as $$\frac{\text{(next after last)} - \text{first}}{\text{ratio} - 1}$$

e.g.

$$2 + 4 + 8 + \ldots + 256 = \frac{512 - 2}{2 - 1}$$

But in a pinch, you can always just rederive the formula since the method is simple:

\begin{align}(2-1) (2 + 4 + 8 + \ldots + 256) &= (4 - 2) + (8 - 4) + (16 - 8) + \ldots + (512 - 256) \\&= 512 - 2 \end{align}