Working with Geometric Series
Introduced| English | Meaning |
|---|---|
| geometric series/ˌdʒiːəʊˈmetrɪk ˈsɪəriːz/ | An infinite series a + ar + ar² + … with constant ratio r; for nonzero a it converges to a/(1 − r) exactly when |r| < 1. |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | The factor multiplying each term to obtain the next term in a geometric sequence. |
| first term | first term |
| partial sum/ˈpɑːʃl sʌm/ | partial sum |
| converges/kənˈvɜːdʒɪz/ | Approaches a finite limit as the relevant variable or sequence index increases or approaches a specified value. |
| diverges/daɪˈvɜːdʒɪz/ | An improper integral, sequence or sequence of series partial sums does not approach a finite limit. |
Repeated halves can have a finite total
- A mathematical model adds $3$ litres, then $1.5$, then $0.75$, always adding half the previous amount. The total approaches $6$ litres.
- This is a geometric series 几何级数. Its terms come from multiplying by a fixed common ratio 公比 $r$.
Identify the actual first term
- Write the series as $a+ar+ar^2+\cdots$. Here $a$ is the first term 首项, not necessarily the number printed before a power.
- For nonzero terms, divide a term by the previous one to find $r$. With $r=0$, later terms are zero; do not try to divide by zero.
In the geometric sum formula a/(1−r), a is the ____ term as written.
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Substitute the starting index to identify the actual first term.
Separate a term from a partial sum
- A partial sum 部分和 adds finitely many terms. The first $N$ terms are $S_N=a+ar+\cdots+ar^{N-1}$.
- In $3+1.5+0.75+\cdots$, the third term is $0.75$, but $S_3=5.25$. Neither number is the infinite sum $6$.
In 3+1.5+0.75+…, what is the third term?
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The term is 0.75; the third partial sum is 5.25.
What is the sum of the first three terms of 3+1.5+0.75+…?
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Add 3,1.5 and 0.75.
Find the finite-sum formula
- Multiply $S_N$ by $r$ and subtract. The middle terms cancel, giving $(1-r)S_N=a-ar^N$.
- Therefore $S_N=a(1-r^N)/(1-r)$ for $r\ne1$. If $r=1$, all $N$ terms equal $a$, so $S_N=Na$.
The upper index is $N-1$ because the first term has index $0$.
In $S_N=\sum_{n=0}^{N-1}ar^n$, what is the upper index for the first four terms?
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The indices are 0,1,2,3.
For r≠1, which is the sum of the first N terms of a+ar+ar²+…?
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Subtract rS_N from S_N to cancel the middle terms.
Check convergence before taking an infinite sum
- For $a\ne0$, the series converges 收敛 exactly when $|r|<1$. Then $r^N\to0$, so $S_N\to a/(1-r)$.
- A ratio can be negative or irrational. Its magnitude decides this test; being written as a fraction is not the condition.
For $3+1.5+0.75+\cdots$, $a=3$ and $r=1/2$:
For the first four terms, $S_4=3(1-(1/2)^4)/(1-1/2)=5.625$. The remaining amount is $6-5.625=0.375$.
Terms, partial sums and the infinite limit
The default series starts at 3 and halves each term. Its partial sums approach 6; an individual term and a partial sum are different quantities. Boundary and zero cases are checked explicitly.
For a nonzero real first term a, $\sum_{n=0}^{\infty}ar^n$ converges exactly when...
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The power r^N tends to 0 exactly when |r|<1.
For |r|<1, what is the infinite sum with first term a and ratio r?
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Take the limit of the finite-sum formula.
For first term 3 and ratio 1/2, what is S₄?
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The first four terms are 3,1.5,0.75 and 0.375.
The infinite sum is 6 and the first four-term sum is 5.625. What remains?
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Subtract the partial sum from the limit.
Keep the exact ratio while calculating
- For $\sum_{n=0}^{\infty}5(1/3)^n$, the first term is $5$ and the ratio is exactly $1/3$. Since $|r|<1$, its sum is $5/(1-1/3)=15/2$.
- Replacing $1/3$ with $0.33$ changes the series. Use the exact value until the final numerical answer, here $7.5$.
Find $\sum_{n=0}^{\infty}5(1/3)^n$.
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$5/(1-1/3)=15/2=7.5$.
Replacing the exact ratio 1/3 with 0.33 leaves a geometric series unchanged.
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Those are different multipliers and give different exact sums.
A negative ratio can still converge
- In $4-2+1-1/2+\cdots$, $r=-1/2$. The partial sums $4,2,3,2.5,\ldots$ approach $4/(1+1/2)=8/3$.
- They alternate around the limit. Oscillation alone does not prove divergence; here the size of each new term decreases to zero.
What is the common ratio in 4−2+1−1/2+…?
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Divide −2 by 4 to get −1/2.
Find the infinite sum of 4−2+1−1/2+….
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The sum is 4/(1+1/2)=8/3.
If partial sums alternate above and below a fixed number, the series must diverge.
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For ratio −1/2 the oscillations shrink and the partial sums converge.
Check the boundaries and the zero case
- With $a\ne0$, $r=1$ gives $S_N=Na$, while $r=-1$ makes the partial sums alternate between $a$ and $0$. Each series diverges 发散. If $|r|>1$, the terms do not tend to zero either.
- If $a=0$, every term is zero for any fixed finite $r$, so the sum is $0$. The multiplier is not uniquely determined by zero terms; at $r=1$, do not use the undefined fraction $0/(1-1)$.
Which series diverges? Each sum begins at n=0.
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The ratio 2 has magnitude above 1 and the nonzero terms do not tend to 0.
The series $\sum_{n=0}^{\infty}0\cdot 2^n$ has sum 0.
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Every term and every partial sum is 0, even though the chosen multiplier is 2.
For a nonzero first term, which ratios give a convergent geometric series?
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All three selected ratios have magnitude below 1; the ratio need not be rational.
A different starting index changes the first term
- For $\sum_{n=1}^{\infty}3\cdot2^{-n}$, substituting $n=1$ gives the first term $3/2$. The ratio is $1/2$, so the sum is $3$.
- For $\sum_{n=2}^{\infty}12(-1/3)^n$, the first term is $4/3$. The ratio is $-1/3$, so the sum is $(4/3)/(1+1/3)=1$.
What is the first term of $\sum_{n=1}^{\infty}3\cdot 2^{-n}$?
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Substitute n=1, giving 3/2.
Find $\sum_{n=2}^{\infty}12(-1/3)^n$.
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The first term is 4/3 and the ratio is −1/3.
Algebra alone does not certify an infinite sum
- For $3+6+12+\cdots$, the terms do not tend to zero. Substituting into $a/(1-r)$ gives $-3$, but that is not an ordinary infinite sum of these positive terms.
- A finite sum is still defined. For example, its first three terms add to $21$. Convergence concerns the limit as the number of terms grows.
For a nonzero first term, apply the infinite-sum formula only after checking $|r|<1$. Read the first term from the starting index. A short plotted pattern supports intuition; it does not replace the convergence argument.
For a nonzero first term, a/(1−r) gives the ordinary infinite sum even when |r|≥1.
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For a nonzero first term the ordinary sum formula requires |r|<1.
Practise new starting values and signs
- Find the first term, ratio and sum when the series begins at $n=1$ with $9(-2/3)^n$.
- Next, begin at $k=3$ with $16(1/2)^k$. Find its infinite sum, first four-term sum and remaining amount.
Find the first term, ratio and sum of $\sum_{n=1}^{\infty}9(-2/3)^n$. Explain convergence.
The starting index changes the first term, not the common ratio.
For $\sum_{k=3}^{\infty}16(1/2)^k$, find the first term, infinite sum, sum of the first four terms and remaining amount.
The first exponent is 3, so the first term is 2.
Compare $\sum_{n=0}^{\infty}5(-1)^n$ and $\sum_{n=0}^{\infty}0\cdot 2^n$. State whether each converges and justify using partial sums.
Separate a nonzero alternating boundary case from the all-zero series.
State what your answer represents
- Identify the first term and ratio, then distinguish an individual term, a finite partial sum and an infinite limit.
- State the convergence reason. For a nonzero first term it is $|r|<1$; handle the all-zero series directly.
For $a\ne0$, $a+ar+ar^2+\cdots$ converges exactly when $|r|<1$, with sum $a/(1-r)$. Its first $N$ terms have sum $a(1-r^N)/(1-r)$ when $r\ne1$. If $a=0$, all terms and every partial sum are zero.
Review: find the actual first term of $\sum_{n=2}^{\infty}20(1/2)^n$.
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Substitute n=2 to get 20/4=5.