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