Geometric sums, alternating ratios and convergence
Введено| English | Значение |
|---|---|
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | знаменатель геометрической прогрессии |
Why does a negative ratio not automatically prevent convergence?
- Alternating payments can have a stable accumulated total.
- This lesson studies common ratio 公比: The fixed factor multiplying each term to obtain the next.
The method
- For first term a and nonzero ratio r, u_n=ar^(n−1).
- If r=0, the terms are a,0,0,…, avoiding a 0^0 expression at the first term.
- For r≠1, the finite sum is S_n=a(1−r^n)/(1−r); for r=1 it is na.
- For nonzero a, an infinite geometric sum exists when |r|<1 and equals a/(1−r).
- A negative ratio alternates signs; convergence depends on its magnitude, not its sign alone.
Why does the series with a=6, r=-1/2 converge despite alternating signs?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
Only the size of the ratio decides: |r|=1/2<1, so the terms tend to zero and the sum is 6/(1+1/2)=4. The sign merely makes partial sums alternate above and below 4.
A worked case
For a=6 and r=−1/2, the terms are 6,−3,3/2,−3/4,… . The first four sum to 15/4=3.75. The finite formula gives S₄=6[1−(−1/2)^4]/(1+1/2)=15/4. Since |r|=1/2<1, S infinity=6/(1+1/2)=4. The signed remainder is 4−S_n=4(−1/2)^n, so odd partial sums are above 4 and even partial sums below 4. For error below 0.01, require 4/2^n<0.01. At n=8 the error is 1/64>0.01; at n=9 it is 1/128<0.01, so 9 terms first meet the condition.
Geometric sums, alternating ratios and convergence
For first term a and nonzero ratio r, u_n=ar^(n−1)
Check which series formula answers the question and whether its conditions hold.
Find u₃ for a=6 and r=−1/2.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
u₃=6(−1/2)²=6/4=1.5.
Where it goes wrong
- Do not replace r=−1/2 by r=1/2 in the finite formula: the parity of n matters.
- The infinite sum formula is not justified at r=1 or r=−1 for nonzero a; their terms do not tend to zero.
- A finite sum can exist even when its infinite continuation diverges.
Not true: Every geometric series with a negative ratio diverges. A geometric series converges when |r|<1, and negative ratios such as −1/2 are included.
Find its first-four-term sum.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
6−3+3/2−3/4=15/4=3.75.
A geometric series converges when |r|<1, and negative ratios such as −1/2 are included.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. A geometric series converges when |r|<1, and negative ratios such as −1/2 are included.
Applying it
- Check the first few terms and partial sums directly before using a formula.
- If |r|>1 and a is nonzero, term magnitudes grow instead of vanishing.
- At r=−1, partial sums alternate between a and 0 rather than approaching one value.
- The zero sequence a=0 is a degenerate exception; identify it rather than applying a nonzero-first-term convergence claim.
Find its infinite sum.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Since |r|<1, the infinite sum is 6/(1+1/2)=4.
Match each geometric quantity to its formula.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The exponent is n-1 because the first term carries r^0; the infinite formula is unjustified at r=±1 where terms do not shrink.
In your course
- AQA A-level Mathematics (7357), specification section D.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Common ratio: The fixed factor multiplying each term to obtain the next.
- For first term a and nonzero ratio r, u_n=ar^(n−1).
- Watch for: do not replace r=−1/2 by r=1/2 in the finite formula: the parity of n matters.