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.
$\endgroup$ 153 Answers
$\begingroup$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 $$\bbox[5px,border:2px solid red]{\sum\limits_{n=N}^\infty ar^n = \frac{ar^N}{1-r}.}$$
$\endgroup$ 2 $\begingroup$$$\sum_{n=N}^\infty ar^n=\sum_{n=0}^\infty ar^{n+N}=r^N\sum_{n=0}^\infty ar^n=r^N\left(\frac{a}{1-r}\right)=\frac{ar^N}{1-r}\ \forall\ |r|<1$$
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.
$\endgroup$ $\begingroup$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:
$$ \begin{align}(2-1) (2 + 4 + 8 + \ldots + 256) &= (4 - 2) + (8 - 4) + (16 - 8) + \ldots + (512 - 256) \\&= 512 - 2 \end{align}$$
$\endgroup$
