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Provided definition, examples, as well as applications for their use and importance.
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Alexander Molnar
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The Coding Aviator
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@ -3,42 +3,146 @@ title: Geometric Series
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---
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## Geometric Series
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A geometric series is a sum of numbers, such as
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```
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1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
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```
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where the first term is not zero and each consecutive pair of terms in the sum has the same ratio, called the *common ratio*. In the example above we have `1/(1/2) = 2`, `(1/2)/(1/4) = 2`, etc... While seemingly fairly simple this type of sum is found extensively in many areas including [geometry of real numbers](https://en.wikipedia.org/wiki/Cantor_set), [fractals](https://en.wikipedia.org/wiki/Geometric_series#Fractal_geometry), [probability](https://math.stackexchange.com/questions/1164163/introduction-to-probability-dice), [economics](https://en.wikipedia.org/wiki/Annuity#Proof_of_annuity-immediate_formula), converting rational numbers from decimal form to fraction form (see below) and more.
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A geometric progression is a sequence of numbers, whose first term is non zero and each term is obtained by multiplying its proceding term by a constant quantity. This constant quantity is called common ratio of the GP.
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We will show how to sum a geometric series when possible, give the formulas and then give some examples of this below.
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The sum of this progression is called geometric series.
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It is common to see the common ratio between terms denoted by `r`, so if a geometric series starts with first term `a`, the series is
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```
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a + a*r + a*r^2 + a*r^3 + a*r^4 + ...
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```
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When summing an infinite number of terms (such as 1 + 1 + 1 + 1 + ...) it is not clear the sum should be a number, but a remarkable property about geometric series is that the sum is always a number when `-1 < r < 1` and otherwise the sum is not a number (e.g., our `1 + 1 + 1 + ...` series tends to infinite, so is not a number). To see why, suppose we are only interested in the first `n + 1` terms of the series,
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```
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a + a*r + a*r^2 + ... + a*r^n
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```
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This is a sum of a `n + 1` numbers, so is certainly a real number itself. If we call this number `S_n`, so
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```
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S_0 = a,
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S_1 = a + a*r,
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S_2 = a + a*r + a*r^2,
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etc...
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```
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we have
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```
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S_n = a + a*r + a*r^2 + ... + a*r^n
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```
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and multiplying both sides by `r` gives
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```
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S_n*r = a*r + a*r^2 + a*r^3 + ... + a*r^n + a*r^(n+1)
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```
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which looks almost like our original sum `S_n` on the right. We have an extra `a*r^(n+1)` and we are missing the first term `a`, so writing
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```
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S_n*r = 0 + a*r + a*r^2 + a*r^n + a*r^(n+1)
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= (a - a) + a*r + a*r^2 + ... + a*r^n + a*r^(n+1)
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= a + a*r + a*r^2 + ... + a*r^n + a*r^(n+1) - a
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```
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we can now see `S_n` on the right hand side again. Indeed, we see that
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```
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S_n*r = S_n + a*r^(n+1) - a
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```
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and
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```
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a - a*r(n+1) = S_n - S_n*r = S_n*(1 - r)
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```
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so that
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```
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S_n = [a - a*r^(n+1)]/(1 - r) = a*[1 - r^(n+1)]/(1 - r).
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```
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when `r` is not equal to 1 (we cannot divide by 0!). Now if `-1 < r < 1` then we know that `r^(n+1)` tends to 0 as `n` tends to infinity, so our sum `S_n` tends to
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```
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S = a*[1 - 0]/(1 - r) = a/(1 - r).
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```
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On the other hand, when `r > 1` or `r < 1` we know that `r^(n_1)` does not tend to a finite number, so the `S_n` do not tend to a finite number either. The only case we have not mentioned yet is when `r = 1`, but then the series is
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```
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S_n = a + a*1 + a*1^2 + ... + a*1^n = a + a + ... + a = (n + 1)*a
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```
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which certainly does not equal a number when `n` tends to infinity.
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We can say if a1,a2,a3,....,an are in GP then
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### Formula for geometric series
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<a href="https://www.codecogs.com/eqnedit.php?latex=\frac{a_{2}}{a_{1}}&space;=&space;\frac{a_{3}}{a_{2}}=...\frac{a_{n}}{a_{n-1}}&space;=&space;r" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\frac{a_{2}}{a_{1}}&space;=&space;\frac{a_{3}}{a_{2}}=...\frac{a_{n}}{a_{n-1}}&space;=&space;r" title="\frac{a_{2}}{a_{1}} = \frac{a_{3}}{a_{2}}=...\frac{a_{n}}{a_{n-1}} = r" /></a>
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If we have some real number `a` and some `r` such that `-1 < r < 1`, then the *finite* geometric series
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```
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a + a*r + a*r^2 + ... + a*r^(n+1)
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```
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has the sum
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```
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a*[1 - r^(n+1)]/(1 - r)
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```
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while the *infinite* geometric series
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```
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a + a*r + a*r^2 + ...
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```
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has the sum
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```
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a/(1 - r).
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```
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<a href="https://www.codecogs.com/eqnedit.php?latex=\boldsymbol{r}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\boldsymbol{r}" title="\boldsymbol{r}" /></a> is known as common ratio of GP.
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if a is the first term then GP can be written as <a href="https://www.codecogs.com/eqnedit.php?latex=a,ar,ar^{2},...,ar^{n-1}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?a,ar,ar^{2},...,ar^{n-1}" title="a,ar,ar^{2},...,ar^{n-1}" /></a>
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### Examples
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example of a GP : 2,4,8,16,....
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For example, the series with `a = 1` and `r = 1/2` above has sum
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```
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1/(1 - 1/2) = 1/(1/2) = 2.
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```
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(This particular sum is well known for [one](https://www.reddit.com/r/Jokes/comments/1423a4/an_infinite_number_of_mathematicians_walk_into_a/) or [two](https://www.reddit.com/r/Jokes/comments/929r9g/an_infinite_number_of_mathematicians_walk_into_a/) jokes.)
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## The nth term of GP (Also known as General term)
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The geometric series with first term `a = 2.5` and common ratio `r = -1/3` has a sum of
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```
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2.5/[1 - (-1/3)] = 2.5/[1 + 1/3] = (5/2)/(4/3) = (5/2)*(3/4) = 15/8
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```
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whereas if we started with `a = -2.5` instead, the series has a sum of
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```
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-2.5/[1 - (-1/3)] = -15/8.
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```
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Let a be the first term, r be the common ratio and l be the last term of a GP, then nth term is given by
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### Application for converting decimals to fractions
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<a href="https://www.codecogs.com/eqnedit.php?latex=T_{n}&space;=&space;l&space;=&space;ar^{^{n-1}}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?T_{n}&space;=&space;l&space;=&space;ar^{^{n-1}}" title="T_{n} = l = ar^{^{n-1}}" /></a>
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where <a href="https://www.codecogs.com/eqnedit.php?latex=r=\frac{T_{n}}{T_{n-1}}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?r=\frac{T_{n}}{T_{n-1}}" title="r=\frac{T_{n}}{T_{n-1}}" /></a>
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# The sum of n terms of a GP
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Let a be the first term, r be the common ratio and l be the last term of a GP, then sum of n terms is given by:
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<a href="https://www.codecogs.com/eqnedit.php?latex=S_{n}=\left\{\begin{matrix}\frac{a\left&space;(&space;r^{n}&space;-1\right&space;)}{r-1},&space;r>1&space;&&space;\\&space;\frac{a\left&space;(&space;1-r^{n}&space;\right&space;)}{1-r},&space;r<1&space;&&space;\end{matrix}\right." target="_blank"><img src="https://latex.codecogs.com/gif.latex?S_{n}=\left\{\begin{matrix}\frac{a\left&space;(&space;r^{n}&space;-1\right&space;)}{r-1},&space;r>1&space;&&space;\\&space;\frac{a\left&space;(&space;1-r^{n}&space;\right&space;)}{1-r},&space;r<1&space;&&space;\end{matrix}\right." title="S_{n}=\left\{\begin{matrix}\frac{a\left ( r^{n} -1\right )}{r-1}, r>1 & \\ \frac{a\left ( 1-r^{n} \right )}{1-r}, r<1 & \end{matrix}\right." /></a>
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and <a href="https://www.codecogs.com/eqnedit.php?latex=S_{n}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?S_{n}" title="S_{n}" /></a> is not defined for r=1
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To see how this can be useful converting decimal numbers to fractions, consider the number `0.1111111...` with 1s repeating. As
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```
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0.1 = 1/10,
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0.01 = 1/100,
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0.001 = 1/1000,
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etc...
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```
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we can rewrite this as
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```
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0.11111... = 1/10 + 1/100 + 1/1000 + ...
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```
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which is a geometric sum with first term `a = 1/10` and common ratio `r = 1/10`, so using our formula for the sum above,
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```
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0.111... = (1/10)/(1 - 1/10) = (1/10)/(9/10) = (1/10)*(10/9) = 1/9.
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```
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# Geometric Mean
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if we insert geometric mean between two numbers <a href="https://www.codecogs.com/eqnedit.php?latex=n_{1}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?n_{1}" title="n_{1}" /></a> and <a href="https://www.codecogs.com/eqnedit.php?latex=n_{2}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?n_{2}" title="n_{2}" /></a> , then
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For a more complicated decimal, consider `0.124312431243...` where the repeating part is 1243. The decimal is equivalent to
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```
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1243/10000 + 1243/10000^2 + 1243/10000^3 + ...
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```
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and so our simple sum formula gives
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```
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0.124312431243... = (1243/10000)/(1 - 1/10000)
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= (1243/10000)*(10000/9999)
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= 1243/9999 = 113/909
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```
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Geometric mean = <a href="https://www.codecogs.com/eqnedit.php?latex=\sqrt{n_{1}*n_{2}}" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\sqrt{n_{1}*n_{2}}" title="\sqrt{n_{1}*n_{2}}" /></a>
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For the most complicated type of repeating decimal, consider one that starts with something different from the repeating pattern, e.g., `0.42567676767...` where after the initial 425 we repeat with 67 forever. This is simply
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```
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0.425 + 0.000676767...
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```
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and so we have
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```
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0.425676767... = 425/1000 + 67/100000 + 67/(100000*100) + 67/(100000*100^2) + ...
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= 425/1000 + (67/100000)/(1 - 1/100)
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= 425/1000 + 67/99000)
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= (425*99 + 67)/99000 = 21071/49500.
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```
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This may be less pleasant without a calculator, but just as simple as the examples above with our wonderful formula!
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This also gives a nice validation that `0.999... = 1`. Indeed, we have
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```
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0.999 = 9/10 + 9/100 + 9/1000 + ...
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= (9/10)/(1 - 1/10)
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= (9/10)/(9/10) = 1.
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```
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