AP Calculus AB/BC · Unit 10 · Topic 10.2 · BC Only
Working with Geometric Series
Recognize constant-ratio structure, derive its partial sums, and evaluate finite sums, infinite sums, and tails.
1. Topic Focus
Determine series convergence, estimate error, construct Taylor approximations, and represent functions with power series on valid intervals.
This topic: Recognize constant-ratio structure, derive its partial sums, and evaluate finite sums, infinite sums, and tails.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Identify the actual first term from the starting index and verify the convergence condition before using the infinite-sum formula.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
BC-only topic. A geometric series is one of the few infinite series in AP Calculus BC whose exact sum can often be found directly. Its special feature is a constant multiplier between consecutive nonzero terms.
Recognize geometric structure
A geometric series can be written as
where \(a\) is the first term and \(r\) is the common ratio. For explicit terms \(u_n\), verify
where the quotient is defined. A constant difference indicates an arithmetic sequence, not a geometric series.
Derive the finite partial-sum formula
For \(r\ne1\), let
Multiplying by \(r\) and subtracting cancels the interior terms:
This is a finite identity and is valid for any real \(r\ne1\). If \(r=1\), then \(S_N=Na\).
Pass from finite to infinite
The infinite series converges precisely when the partial sums approach a finite limit. Since \(r^N\to0\) exactly when \(|r|<1\),
The condition is part of the conclusion, not an optional check.
What happens when \(|r|\ge1\)?
- If \(r=1\) and \(a\ne0\), partial sums grow linearly.
- If \(r=-1\) and \(a\ne0\), partial sums oscillate.
- If \(|r|>1\), the terms do not approach zero and their magnitudes grow.
Except for the trivial case \(a=0\), the series diverges whenever \(|r|\ge1\).
Starting index determines the first term
In a sigma expression, substitute the lower index before using \(a/(1-r)\):
The coefficient \(c\) is not necessarily the first term. The first term is \(cr^m\).
Negative ratios
When \(-1<r<0\), terms alternate signs while shrinking in magnitude. Partial sums usually approach the sum from alternating sides. Keep the signed ratio in the denominator:
Tails and exact remainders
For \(\sum_{n=0}^{\infty}ar^n\), the tail beginning with exponent \(m\) is
If \(S_N=\sum_{n=0}^{N}ar^n\), the exact remainder after \(S_N\) is
This formula preserves the sign of an alternating remainder. Its absolute value gives the error magnitude.
Reindex without changing terms
The same series may appear as
When reindexing, adjust both the exponent and limits so the ordered list of terms remains unchanged.
Repeating decimals
A repeating block produces place values in a geometric pattern. For example, a two-digit repeating block has ratio \(10^{-2}\). Express the decimal as a series and sum it exactly rather than relying on rounded calculator output.
Repeated-process models
Rebounds, repeated reflections, retained medication, and recurring deposits can create geometric contributions. Define what one term measures, identify whether an initial quantity is counted once or twice, and check that a constant ratio is reasonable before summing.
Bridge to power series
The identity
will later generate many power-series representations through substitution, multiplication, differentiation, and integration.
Geometric-series checklist
- Write several terms and verify a constant ratio.
- Identify the actual first term from the lower index.
- Distinguish a finite sum from an infinite sum.
- For an infinite series, state and verify \(|r|<1\).
- Use the signed ratio in \(a/(1-r)\).
- Check the answer against the signs and approximate size of the first few partial sums.
6. Detailed Worked Example and Error Check
Example 1: An alternating geometric series
The series
has first term \(a=5\) and ratio \(r=-1/3\). Since \(|r|<1\),
Example 2: Starting index is not zero
For
the first term is \(3(1/2)^2=3/4\), not \(3\). Therefore
Example 3: Ratio \(-1\)
The series \(4-4+4-4+\cdots\) has \(r=-1\). Its partial sums alternate between \(4\) and \(0\), so it diverges. The expression \(4/[1-(-1)]\) is invalid because \(|r|<1\) is not satisfied.
Example 4: Terms grow in magnitude
For \(\sum_{n=0}^{\infty}2(3/2)^n\), the ratio is \(3/2\). The terms do not approach zero, so the series diverges.
Example 5: A finite geometric sum
Evaluate \(\sum_{n=0}^{7}2(1/3)^n\). There are eight terms:
No convergence condition is needed for a finite sum.
Example 6: Sum only a tail
For \(\sum_{n=5}^{\infty}7(0.2)^n\), the first included term is \(7(0.2)^5=7/3125\). Thus
Example 7: Convert a repeating decimal
The decimal \(0.\overline{27}\) can be written
Here \(a=27/100\) and \(r=1/100\), so
Example 8: Total distance of a bouncing ball
A ball is dropped \(10\) meters and rebounds to \(60\%\) of each previous height. The initial drop occurs once; every rebound height is traveled upward and downward:
Example 9: Repeated contributions
A process contributes \(200\) units initially, then \(40\%\) of the preceding contribution each stage:
The model converges because each new contribution is a fixed fraction less than one in magnitude.
Example 10: Determine a parameter from a sum
Suppose
Under the required condition \(|x-1|<1\),
The result is valid because \(|7/4-1|=3/4<1\).
Common errors
- Using a difference instead of a ratio to identify the series.
- Calling the coefficient outside the power the first term without checking the lower index.
- Using \(a/(1-r)\) when \(|r|\ge1\).
- Replacing a negative ratio by its absolute value in the denominator.
- Using the infinite-sum formula for a finite series.
- Counting \(N\) terms in a sum from exponent \(0\) through exponent \(N\), which has \(N+1\) terms.
- Starting a tail at \(r^{m-1}\) instead of its actual first exponent \(r^m\).
- Double-counting or omitting the initial movement in a repeated-distance model.
- Reindexing the exponent without changing the bounds.
- Solving for a parameter but failing to verify \(|r|<1\).
7. AP Reasoning Routine
Check the nth-term condition first, match the series structure to a justified test, state convergence type, and test power-series endpoints separately.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Classify or evaluate each series, showing the first term and ratio.
(a) Evaluate \(4-2+1-\tfrac12+\cdots\).
(b) Evaluate \(\sum_{n=3}^{\infty}5(1/4)^n\).
(c) Determine whether \(\sum_{n=1}^{\infty}6(-2)^{n-1}\) converges.
(d) Evaluate the finite sum \(\sum_{n=1}^{6}3(2/3)^{n-1}\).
(e) For \(\sum_{n=0}^{\infty}ar^n\), write the exact remainder after \(S_N=\sum_{n=0}^{N}ar^n\).
(f) Find the exact error after four terms of \(1+\tfrac13+\tfrac19+\cdots\), where the four terms have exponents \(0\) through \(3\).
(g) Express \(0.\overline{18}\) as a fraction using a geometric series.
(h) A ball is dropped \(12\) meters and rebounds to half its previous height. Find its total vertical distance.
(i) Find \(x\) if \(\sum_{n=0}^{\infty}x^n=5\), and verify convergence.
(j) A series begins \(12,-3,3/4,-3/16,\ldots\). Identify \(a\) and \(r\), then find its sum.
(k) Explain why changing the first three terms of a convergent geometric series does not change convergence but generally changes its sum.
(l) Immediately after a dose, a medication adds \(100\) mg. Just before each later dose, \(70\%\) of the preceding amount remains. Model the long-run amount immediately after dosing as a geometric series and find its value.
Check the solution
(a) \(a=4\), \(r=-1/2\), so the sum is \(4/[1-(-1/2)]=8/3\).
(b) The first term is \(5(1/4)^3=5/64\), so the sum is \((5/64)/(3/4)=5/48\).
(c) The ratio is \(-2\), whose magnitude exceeds \(1\). The terms do not approach zero, so the series diverges.
(d) There are six terms: \(S_6=3[1-(2/3)^6]/(1-2/3)=665/81\).
(e) \(R_N=ar^{N+1}/(1-r)\), valid when \(|r|<1\).
(f) The first omitted term is \((1/3)^4\), so \(R_3=(1/3)^4/(1-1/3)=1/54\).
(g) \(0.\overline{18}=0.18+0.0018+\cdots=(18/100)/(1-1/100)=18/99=2/11\).
(h) \(D=12+2\sum_{n=1}^{\infty}12(1/2)^n=12+24=36\) meters.
(i) \(1/(1-x)=5\), so \(x=4/5\). Since \(|4/5|<1\), the series converges.
(j) \(a=12\) and \(r=-1/4\). The sum is \(12/[1-(-1/4)]=48/5\).
(k) Convergence depends on the infinite tail and is unaffected by finitely many terms. The sum changes by the net amount of the finite modifications.
(l) The post-dose contributions are \(100+70+49+\cdots\), so the limiting amount is \(100/(1-0.7)=1000/3\) mg.