Find a finite or infinite sum
Subtract a shifted copy
The top row shows the sum \(S=80+40+20+10+5\), and each bar is half as tall as the bar before it.
Press the first button. The middle row shows half of \(S\) moved one place to the right, which is \(\frac12S=40+20+10+5+2.5\).
Press the second button to subtract the middle row from the top row. The bars 40, 20, 10 and 5 cancel, so only the teal bar 80 and the red bar \(-2.5\) are left. The red bar lies below the axis because it is subtracted, and \(S-\frac12S=80-2.5=77.5\).
Then \(\frac12S=77.5\), so \(S=155\). As a check, \(80+40+20+10+5=155\).
Move the slider to 10 terms. The red leftover bar shrinks toward 0, so if the bars continue forever, \(\frac12S=80\) and \(S=160\).
The first term over one minus the ratio
Without the numbers, the top row shows \(S_n=a+ar+\cdots+ar^{n-1}\), a sum of \(n\) terms. The middle row shows \(rS_n=ar+ar^2+\cdots+ar^n\), the same bars times \(r\) and moved one place to the right.
Subtract the middle row from the top row. The middle terms cancel, so \((1-r)S_n=a-ar^n\).
If \(r\ne1\), divide by \(1-r\). The sum of \(n\) terms is \(S_n=\frac{a(1-r^n)}{1-r}\).
The red bar is the leftover term \(ar^n\), which is subtracted. If \(|r|<1\), it shrinks toward 0 as \(n\) grows, and then the sum of all the terms is \(\frac{a}{1-r}\).
Notation
The sum of \(n\) terms, with \(r\ne1\), uses the first term \(a\) and the number of terms \(n\).
\[S_n=\cy{a}\cdot\dfrac{1-r^{\,n}}{1-r}\]The letter \(n\) is the number of terms, not the last exponent. If the index runs from \(p\) to \(q\), then \(n=q-p+1\).
If \(r=1\), every term is \(a\), so the sum of \(n\) terms is \(na\).
For infinitely many terms, the sum is the first term divided by one minus the ratio. This needs \(|r|<1\). Here \(a\) is the first term of the sum as written, whatever its index.
\[\cy{a}+\cy{a}r+\cy{a}r^2+\cdots=\dfrac{\cy{a}}{1-r}\]
When it fails
Set \(r=2\) with \(a=10\). The formula \(\frac{a}{1-r}\) gives \(\frac{10}{1-2}=-10\), the red dashed line. But every term is positive, so no sum of these terms can be negative.
If \(|r|\ge1\), the leftover term \(ar^n\) does not shrink toward 0, so the sums do not approach \(\frac{a}{1-r}\). Use \(\frac{a}{1-r}\) only when \(|r|<1\).
Example 1: An infinite sum with a negative ratio
Find the sum of \(27-18+12-8+\cdots\).
The first term is 27. The red bars below the axis are the negative terms, so the ratio is negative. It is the second term divided by the first.
\[a=27,\qquad r=\frac{-18}{27}=-\frac23\]Because \(|r|=\frac23\) is less than 1, the formula for an infinite sum applies.
\[S=\frac{a}{1-r}=\frac{27}{1+\frac23}\]Simplify the denominator: \(1+\frac23=\frac53\). Then divide. The dots for the partial sums swing above and below the dashed line at this sum, closer each time.
\[S=27\cdot\frac35=\frac{81}{5}\]
Example 2: Counting the terms
Find \(\displaystyle\sum_{k=2}^{7}3\cdot2^k\).
The first term is the term at \(k=2\), which is \(3\cdot2^2=12\). Each term is 2 times the one before, so the ratio is 2.
A tempting wrong move: The last index is 7, so there are 7 terms.
\[S=12\cdot\frac{1-2^7}{1-2}=1524\]The figure shows 6 bars, for \(k=2,3,4,5,6,7\). A sum of 7 terms adds one more bar, the red outlined bar for \(k=8\), which is 768 and is not in the sum.
Count the terms from the first index to the last: \(7-2+1=6\). Use \(n=6\).
\[S=12\cdot\frac{1-2^6}{1-2}=756\]As a check, add the six terms.
\[\begin{aligned}12+24+48+96&+192+384\\ &=756\end{aligned}\]
Practice
Find the value of \(x\) for which \(6+6x+6x^2+\cdots=15\).
The sum is the first term divided by one minus the ratio. Move the slider until the dashed line meets the orange line at 15.
Solution
The series is geometric with \(a=6\) and \(r=x\). If \(|x|<1\), its sum is \(\frac{6}{1-x}\). Set this equal to 15.
\[\frac{6}{1-x}=15\]Multiply both sides by \(1-x\). Then divide both sides by 15.
\[1-x=\frac{6}{15}=\frac25,\qquad x=\frac35\]The value \(\frac35\) has absolute value less than 1, so the formula applies.
Questions
Questions 1 and 2 use the figure. It shows the first five terms of a geometric series. The heights of the bars are 54, 18, 6, 2 and \(\frac23\).
Which expression gives the total height of the five bars?
- \(\dfrac{54\left(1-(1/3)^5\right)}{1-1/3}\)
- \(\dfrac{54\left(1-(1/3)^4\right)}{1-1/3}\)
- \(\dfrac{54\left(1-(1/3)^6\right)}{1-1/3}\)
Answer
\(\dfrac{54\left(1-(1/3)^5\right)}{1-1/3}\)
If the bars continue forever, what total do they approach?
- 81
- \(\dfrac{242}{3}\)
- 162
Answer
81
What is \(\displaystyle\sum_{n=2}^{\infty}10\left(\frac15\right)^n\)?
- \(\dfrac12\)
- \(\dfrac{25}{2}\)
- \(\dfrac25\)
Answer
\(\dfrac12\)
Summary
You find the sum of a geometric series from its first term, its ratio and, for a finite sum, the number of terms.
Subtracting \(r\) times the sum from the sum cancels all terms but two, and the leftover term shrinks toward 0 when \(|r|<1\).
You use the formula when the ratio is constant, after you count the terms and read the first term from the sum as written.
This is one of four skills in topic 10.2 of Which & Why: AP Calculus.