Geometric sums, alternating ratios and convergence
| English | Español |
|---|---|
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | razón común |
Alternating payments can have a stable accumulated total. Why does a negative ratio not automatically prevent convergence?
- Alternating payments can have a stable accumulated total. Why does a negative ratio not automatically prevent convergence?
- This lesson studies common ratio 公比: The fixed factor multiplying each term to obtain the next.
Choose the mathematical structure
- 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.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines common ratio?
The fixed factor multiplying each term to obtain the next.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
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.
u₃=6(−1/2)²=6/4=1.5.
Test a tempting shortcut
- 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.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every geometric series with a negative ratio diverges. This claim is false. Explain which definition or assumption it violates.
Find its first-four-term sum.
6−3+3/2−3/4=15/4=3.75.
Every geometric series with a negative ratio diverges.
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.
Interpret a new situation
- 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.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find its infinite sum.
Since |r|<1, the infinite sum is 6/(1+1/2)=4.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 7357 · A-level · D. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The fixed factor multiplying each term to obtain the next. Choose the relationship, show the method, check its assumptions and interpret the result.