How to find the sum of a geometric series
When a geometric series converges
We already know from the last section that the standard form of a geometric series is
???\sum^{\infty}_{n=1}ar^{n-1}???
or
???\sum^{\infty}_{n=0}ar^n???
Given either of these forms, the geometric series test for convergence says that
if ???|r|<1??? then the series converges
if ???|r|\ge1??? then the series diverges
When a geometric series converges, we can find its sum.
Sum of a geometric series
We can use the values of ???a??? and ???r??? and the formula for the sum of a geometric series
???\sum^{\infty}_{n=1}ar^{n-1}=\frac{a}{1-r}???
or
???\sum^{\infty}_{n=0}ar^n=\frac{a}{1-r}???
to find the sum of the geometric series.
Finding the sum of a geometric series
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Calculating the sum of a geometric series
Example
Calculate the sum of the geometric series.
???\sum^{\infty}_{n=0}\frac{2^{n-1}}{3^n}???
We showed in the last section that this series was geometric by rewriting it as
???\sum^{\infty}_{n=0}\frac{2^{n-1}}{3^n}???
???\sum^{\infty}_{n=0}\frac{2^n2^{-1}}{3^n}???
???\sum^{\infty}_{n=0}2^{-1}\left(\frac{2^n}{3^n}\right)???
???\sum^{\infty}_{n=0}\frac12\left(\frac23\right)^n???
Now that we have the series in the right form, we can say
???\sum^{\infty}_{n=0}ar^n=\sum^{\infty}_{n=0}\frac12\left(\frac23\right)^n??? where
???a=\frac12???
???r=\frac23???
Since the sum of a geometric series is given by
???\sum^{\infty}_{n=0}ar^n=\frac{a}{1-r}???
we can say that the sum is
???\sum^{\infty}_{n=0}\frac12\left(\frac23\right)^n=\frac{\frac12}{1-\frac23}???
???\frac{\frac12}{\frac33-\frac23}???
???\frac{\frac12}{\frac13}???
???\frac12\cdot\frac31???
???\frac32???
We could have also found the sum by expanding the series through its first few terms and identifying values for ???a??? and ???r???.
???\sum^{\infty}_{n=0}\frac12\left(\frac23\right)^n=\frac12\left[\left(\frac23\right)^0+\left(\frac23\right)^1+\left(\frac23\right)^2+\left(\frac23\right)^3+...\right]???
???\sum^{\infty}_{n=0}\frac12\left(\frac23\right)^n=\frac12\left(1+\frac23+\frac49+\frac8{27}+...\right)???
So
???a=\frac12???
???r=\frac23???
and
???\sum^{\infty}_{n=0}\frac12\left(\frac23\right)^n=\frac{\frac12}{1-\frac23}???
???\frac{\frac12}{\frac33-\frac23}???
???\frac{\frac12}{\frac13}???
???\frac12\cdot\frac31???
???\frac32???