3.4 Geometric series

A geometric series is anything which looks like

a+ar+ar2++arn1a+ar+ar^{2}+...+ar^{n-1} (3.42)

Here, aa is the starting term, rr is called the \saycommon ratio (this name might make a bit more sense if you consider how each term is rr times smaller than the next term) and nn is the total number of terms. We can write the series in \sum-notation as (it’s worth expanding this and checking that it is indeed the same as Equation 3.42)

j=1n[arj1]\sum_{j=1}^{n}\left[ar^{j-1}\right] (3.43)

3.4.1 Finding the sum of a geometric series

This is one of those things which is much harder to find than it is to verify that one has indeed found the correct thing. For A Level maths the sum is given in the formula booklet, and it’s not necessary to be able to show why it is true.

If we call the value of Equation 3.42 SnS_{n} (read as "the sum of nn terms of the series), then we can multiply SnS_{n} by rr and get that

rSn=ar+ar2++arnrS_{n}=ar+ar^{2}+...+ar^{n} (3.44)

From here we can proceed by finding the value of SnrSnS_{n}-rS_{n}

SnrSn\displaystyle S_{n}-rS_{n} =(a+ar+ar2++arn1)(ar+ar2++arn)\displaystyle=(a+ar+ar^{2}+...+ar^{n-1})-(ar+ar^{2}+...+ar^{n}) (3.45)
=aarn\displaystyle=a-ar^{n} (3.46)

This can then be rearranged a bit by factoring out the SnS_{n} and then dividing through by 1r1-r.

Sn(1n)=aarn\displaystyle S_{n}(1-n)=a-ar^{n} (3.47)
Sn=aarn1r\displaystyle S_{n}=\frac{a-ar^{n}}{1-r} (3.48)
Sn=a(1rn)1r\displaystyle S_{n}=\frac{a(1-r^{n})}{1-r} (3.49)

3.4.2 Sum to infinity

The sum to infinity (which I think is in the formula booklet) doesn’t always converge to a specific value (because the terms get bigger and bigger). For example, the value of

1+2+4+8+1+2+4+8+... (3.50)

is infinite, but the value of

1+12+14+18+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... (3.51)

is not infinite, and can be found. If the size (i.e. it doesn’t matter if the value is positive or negative) of the common ratio is less than one, then the terms in the series become successively smaller and smaller (note that just because the size of the terms in a series decreases does NOT mean that it converges to a fixed value, for example the series 1+12++1n+1+\frac{1}{2}+...+\frac{1}{n}+... does not converge). This is because multiplying two decimal numbers together makes a smaller number 44 4 e.g. 1414=116\frac{1}{4}\cdot\frac{1}{4}=\frac{1}{16} and 116\frac{1}{16} is smaller than 14\frac{1}{4}.

If |r|<1\left\lvert r\right\rvert<1 (i.e. if rr is greater than 1-1 and less than 11), then as nn tends to infinity (written nn\to\infty) then rnr^{n} tends to 0 (written rn0r^{n}\to 0). We can apply this to Equation 3.49 and obtain that

S=a1rS_{\infty}=\frac{a}{1-r} (3.52)

3.4.3 "Hidden" geometric series.

Sometimes geometric series can be hiding in plain sight!

For example, this series (note: logarithms and exponents are explored in Section 8) is actually a geometric series!

log3(31312314)\log_{3}\left(3^{1}\cdot 3^{\frac{1}{2}}\cdot 3^{\frac{1}{4}}\cdot...\right) (3.53)

Applying the log rules, we get

log3(31)+log3(312)+log3(314)+\log_{3}\left(3^{1}\right)+\log_{3}\left(3^{\frac{1}{2}}\right)+\log_{3}\left(% 3^{\frac{1}{4}}\right)+... (3.54)

Which simplifies to

1+12+14+1+\frac{1}{2}+\frac{1}{4}+... (3.55)

Which is just an infinite geometric series that converges to 22 (as |r|<1\left\lvert r\right\rvert<1).