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Calculus BC · Unit 10 · 10.2
AP Calculus BC · Infinite Sequences and Series
Working with Geometric Series
The one family whose exact sum is always available — if the ratio is smaller than one.
key idea$\sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}\quad\text{if }\lvert r\rvert\lt 1$
The test
Applies to
$\displaystyle\sum ar^{n}$ — a constant ratio between consecutive terms
Converges when
$\lvert r\rvert\lt 1$
Diverges when
$\lvert r\rvert\ge 1$, including $r=\pm1$
Sum
$\dfrac{\text{first term}}{1-r}$
Common error
using $\dfrac{a}{1-r}$ with $a$ taken as the coefficient rather than the first term actually written
Geometric series are the one family whose exact sum is always available. Identify the ratio $r$ by dividing any term by the one before it; if the result is a constant, the series is geometric and the whole question reduces to comparing $\lvert r\rvert$ with $1$.
When to reach for it
| Series | $r$ | Converges? | Sum |
|---|
| $\displaystyle\sum_{n=0}^{\infty}\frac{1}{2^{n}}$ | $\tfrac12$ | yes | $\dfrac{1}{1-\tfrac12}=2$ |
| $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^{n}}$ | $\tfrac12$ | yes | $\dfrac{\tfrac12}{1-\tfrac12}=1$ |
| $\displaystyle\sum_{n=0}^{\infty}3\left(\tfrac{-2}{3}\right)^{n}$ | $-\tfrac23$ | yes | $\dfrac{3}{1+\tfrac23}=\dfrac95$ |
| $\displaystyle\sum_{n=0}^{\infty}\left(\tfrac54\right)^{n}$ | $\tfrac54$ | no | — |
| $\displaystyle\sum_{n=0}^{\infty}(-1)^{n}$ | $-1$ | no | — |
Rows 1 and 2 differ only in the starting index, and their sums differ accordingly. The numerator of the formula is the first term you actually write down, not the coefficient in front of $r^{n}$.
four facts
- Test for geometric by dividing consecutive terms. A constant quotient is the signature.
- Convergence depends on $\lvert r\rvert$ alone — the first term never affects it.
- The sum is $\dfrac{\text{first term}}{1-r}$, so re-indexing changes the sum but not the verdict.
- $r=1$ and $r=-1$ both diverge, the first growing without bound and the second oscillating.
Worked examples
EX 1
Standard form
read $a$ and $r$ directlyEvaluate $\displaystyle\sum_{n=0}^{\infty}5\left(\dfrac{2}{3}\right)^{n}$.
1
Identify the ratio
Consecutive terms differ by a factor of $\dfrac23$, so $r=\dfrac23$.
2
Check convergence
$\left\lvert\dfrac23\right\rvert\lt 1$, so the series converges.
3
Find the first term
At $n=0$: $5\left(\tfrac23\right)^{0}=5$.
4
Apply the formula
$\dfrac{5}{1-\tfrac23}=\dfrac{5}{\tfrac13}$.
EX 2
A shifted starting index
the first term is not the coefficientEvaluate $\displaystyle\sum_{n=2}^{\infty}\dfrac{4}{3^{n}}$.
1
Identify the ratio
$r=\dfrac13$, so the series converges.
2
Write the first term explicitly
At $n=2$: $\dfrac{4}{9}$.
3
Apply the formula with that first term
$\dfrac{4/9}{1-\tfrac13}=\dfrac{4/9}{2/3}$.
4
Simplify
$\dfrac49\cdot\dfrac32=\dfrac23$.
5
See the trap
Using $4$ as the numerator would give $6$ — the sum from $n=0$, which includes two terms the problem never asked for.
EX 3
Hidden geometric form
rewrite firstEvaluate $\displaystyle\sum_{n=1}^{\infty}\dfrac{2^{n+1}}{5^{n}}$.
1
Split the exponent
$2^{n+1}=2\cdot2^{n}$, so the term is $2\left(\dfrac{2}{5}\right)^{n}$.
2
Identify the ratio
$r=\dfrac25$, and $\left\lvert\dfrac25\right\rvert\lt 1$.
3
Find the first term
At $n=1$: $2\cdot\dfrac25=\dfrac45$.
4
Apply the formula
$\dfrac{4/5}{1-\tfrac25}=\dfrac{4/5}{3/5}$.
5
Evaluate
$\dfrac43$. Splitting the exponent turned an unfamiliar expression into standard form.
EX 4
A repeating decimal
geometric in disguiseExpress $0.\overline{27}$ as a fraction using a geometric series.
1
Write it as a sum
$0.272727\ldots=\dfrac{27}{100}+\dfrac{27}{10000}+\cdots$
2
Identify $a$ and $r$
First term $\dfrac{27}{100}$, ratio $\dfrac{1}{100}$.
3
Check convergence
$\left\lvert\dfrac{1}{100}\right\rvert\lt 1$. ✓
4
Sum it
$\dfrac{27/100}{1-1/100}=\dfrac{27/100}{99/100}=\dfrac{27}{99}$.
5
Simplify
$\dfrac{3}{11}$. Every repeating decimal is a convergent geometric series, which is precisely why it equals a rational number.
EX 5
Solving for the ratio
work backwards from the sumA geometric series with first term $6$ sums to $10$. Find $r$.
1
Write the formula
$\dfrac{6}{1-r}=10$.
2
Clear the fraction
$6=10(1-r)$.
3
Solve
$6=10-10r$, so $10r=4$ and $r=\dfrac25$.
4
Verify convergence
$\left\lvert\dfrac25\right\rvert\lt 1$, so a sum genuinely exists. ✓
5
Note the necessary check
Had the algebra produced $\lvert r\rvert\ge1$, the premise would be contradictory — no such convergent series exists.
What it cannot tell you
The numerator is the first term written
Re-indexing from $n=0$ to $n=2$ changes the sum. Evaluate the term at the actual starting index and use that.
Only a constant ratio counts
$\sum\dfrac{1}{n^{2}}$ has a shrinking ratio that is not constant, so it is not geometric and this formula does not apply to it.
$\lvert r\rvert=1$ diverges
Both $r=1$ and $r=-1$ fail. The formula would divide by zero, which is the algebra warning you.
Convergence ignores the first term
Multiplying every term by $100$ changes the sum by a factor of $100$ and changes the verdict not at all.