Define convergence through partial sums and reject misleading termwise intuition.
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Try
Add 1/2, then 1/4, then 1/8. The terms get smaller while the running total grows. Predict the limit of each sequence separately. Confusing these two limits is one of the most common errors in interpreting an infinite sum.
Predict
If added terms approach zero, must the total settle to a finite number? Compare adding successive powers of 1/2 with adding 1/n. Both term sequences approach zero. Their accumulated behavior requires more than that necessary condition.
Experiment
Construct a two-column ledger for aₙ=1/2ⁿ.
n
Term aₙ
Partial sum Sₙ
1
1/2
1/2
2
1/4
3/4
3
1/8
7/8
4
1/16
15/16
Draw a unit-length bar and shade half the remaining unshaded portion at each step. The shaded total is Sₙ and the latest piece is aₙ. Use the finite identity Sₙ=1−1/2ⁿ to explain why the total approaches 1 while the latest piece approaches zero.
Understand
The infinite series Σₙ₌₁∞aₙ is defined through its partial sums Sₙ=a₁+⋯+aₙ. It converges to S if this sequence has the finite limit S. Divergence can mean unbounded growth, oscillation or another failure of a finite limit. The symbol infinity does not instruct ordinary arithmetic to add a final term at an infinite index.
For n≥2, aₙ=Sₙ−Sₙ₋₁. If the series converges, both partial-sum sequences approach S, so aₙ approaches zero. This is a necessary condition. Its contrapositive is the useful divergence test: if terms fail to approach zero, the series cannot converge. Reversing the implication is invalid.
The harmonic series demonstrates the failure of that converse. Group its terms after the first into blocks ending at powers of two. From index 2^(k−1)+1 through 2^k there are 2^(k−1) terms, each at least 1/2^k. Every block contributes at least 1/2. Infinitely many blocks force unbounded partial sums, even though each individual term tends to zero. A slowly flattening finite plot cannot negate this argument.
Partial-sum formulas can also reveal terms. If Sₙ=2−1/(n+1), then for n≥2 the term is 1/n−1/(n+1). The first term is S₁=3/2, because there is no prescribed S₀ from that formula unless the problem defines it consistently. Carefully checking the starting index prevents silently changing the series. Changing finitely many terms can change its sum but not whether its tail converges.
Numerical computation has practical limits. A running total may stop changing in finite precision because a term becomes too small relative to the stored total, not because the mathematical tail is zero. Conversely, slow divergence can look stable over many iterations. Exact identities, comparison bounds and convergence theorems provide the missing tail information.
Apply
In the practice bank, name the sequence whose limit is requested. Show a finite-sum calculation before taking a limit. Prove the necessary term condition using consecutive partial sums, and use the harmonic grouping argument to explain why that condition is not sufficient.
Continue
Next derive two especially useful finite-sum identities: geometric sums and telescoping sums. Their cancellation patterns make both the accumulated value and the remaining tail visible.
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Practice and reasoning
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Find S₃ for aₙ=1/2ⁿ with indexing beginning n=1.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
If partial sums are Sₙ=2−1/(n+1), find a₃=S₃−S₂.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
If aₙ→0, what can be concluded about Σaₙ without more information?
Explain the difference between the limits of aₙ and Sₙ=Σₖ₌₁ⁿaₖ, using aₙ=1/2ⁿ.
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Prove that convergence of Σaₙ implies aₙ→0.
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A plot of the first 100 harmonic partial sums looks slowly varying. Why is this not a convergence proof?
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Reflect and continue
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