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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.

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  • $\begingroup$ 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. $\endgroup$ – Mathematician 42 Nov 7 '16 at 0:29
  • $\begingroup$ @Mathematician42 Can you please elaborate? I have the formula $\dfrac{a-ar^n}{1-r}$. $\endgroup$ – The Pointer Nov 7 '16 at 0:30
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    $\begingroup$ If $|r|<1$, then $\sum_{n\geq 0} ar^n=a\sum_{n\geq 0}r^n=\frac{a}{1-r}$. Now notice that $\sum_{n\geq m}ar^n=ar^m\sum_{n\geq 0}r^{n}$. $\endgroup$ – Mathematician 42 Nov 7 '16 at 0:32
  • $\begingroup$ @Mathematician42 I do not understand the second part of your explanation. $\endgroup$ – The Pointer Nov 7 '16 at 0:35
  • $\begingroup$ 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. $\endgroup$ – Mathematician 42 Nov 7 '16 at 0:39
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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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    $\begingroup$ Exactly correct. That being said, I find it much much easier to remember as $\frac{\text{first term}}{1-\text{common ratio}}$ $\endgroup$ – erfink Nov 7 '16 at 1:06
  • $\begingroup$ Great explanation. Thank you very much. $\endgroup$ – The Pointer Nov 7 '16 at 1:13
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$$\sum_{n=N}^\infty ar^n=\sum_{n=0}^\infty ar^{n+N}=r^N\sum_{n=0}^\infty ar^n=r^N\left(\frac{a}{1-r}\right)=\frac{ar^N}{1-r}\ \forall\ |r|<1$$

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.

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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:

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

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