riaology
Unit 4 · Lesson 7 of 9

Cancellation can produce conditional convergence.

Positive and negative contributions can settle even when the sum of their magnitudes grows without bound.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its finite partial sum.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase term count n? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Read the model∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

Term count versus Partial sum. Partial sums, Limit ln2. Selected finite partial sum: 0.63452 . Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.000.520140Partial sumTerm count
Partial sumsLimit ln2 · dashed

finite partial sumS8=0.634524,∣R8∣≤19=0.111111S_8=0.634524,\quad|R_8|\le\frac1{9}=0.111111

Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.

Assumptions, units, and limits

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

The alternating-series test requires magnitudes decreasing to zero. Absolute convergence is stronger and implies ordinary convergence. For this alternating harmonic series the next omitted magnitude bounds the remainder, while the absolute harmonic series diverges.

∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}
02

Keep the reasoning visible

∣S−SN∣≤bN+1|S-S_N|\le b_{N+1}

Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.

03

What makes this model useful?

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

Alternating signs alone do not prove convergence. Remainder bounds need the monotonicity and vanishing-term hypotheses.

A worked example

Calculate the starting scenario

  1. Use term count n = 8 .
  2. S8=0.634524,∣R8∣≤19=0.111111S_8=0.634524,\quad|R_8|\le\frac1{9}=0.111111
  3. Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero. The checked quantity is finite partial sum.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the finite partial sum using term count n 8 . State the meaning of its sign and units.

Checked quantity: finite partial sum · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The alternating-series test requires magnitudes decreasing to zero. Absolute convergence is stronger and implies ordinary convergence. For this alternating harmonic series the next omitted magnitude bounds the remainder, while the absolute harmonic series diverges.

Explore the visual solution
Read the model∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

Term count versus Partial sum. Partial sums, Limit ln2. Selected finite partial sum: 0.63452 . Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.000.520140Partial sumTerm count
Partial sumsLimit ln2 · dashed

finite partial sumS8=0.634524,∣R8∣≤19=0.111111S_8=0.634524,\quad|R_8|\le\frac1{9}=0.111111

Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.

Assumptions, units, and limits

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

The alternating-series test requires magnitudes decreasing to zero. Absolute convergence is stronger and implies ordinary convergence. For this alternating harmonic series the next omitted magnitude bounds the remainder, while the absolute harmonic series diverges.

∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}
Calculate and explain

Apply the idea to a new scenario

Use N=15N=15 terms of the alternating harmonic series. Find the partial sum and next-term error bound.

Checked quantity: finite partial sum · dimensionless. Follow the definition for finite-cutoff and numerical-model questions.

Hint

The alternating-series test requires magnitudes decreasing to zero. Absolute convergence is stronger and implies ordinary convergence. For this alternating harmonic series the next omitted magnitude bounds the remainder, while the absolute harmonic series diverges.

Explore the visual solution
Read the model∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

Term count versus Partial sum. Partial sums, Limit ln2. Selected finite partial sum: 0.72537 . Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.000.520140Partial sumTerm count
Partial sumsLimit ln2 · dashed

finite partial sumS15=0.725372,∣R15∣≤116=0.0625S_15=0.725372,\quad|R_15|\le\frac1{16}=0.0625

Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.

Assumptions, units, and limits

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

The alternating-series test requires magnitudes decreasing to zero. Absolute convergence is stronger and implies ordinary convergence. For this alternating harmonic series the next omitted magnitude bounds the remainder, while the absolute harmonic series diverges.

∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Read the model∑n=1∞(−1)n+1n=ln⁡2,∣RN∣≤1N+1\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=\ln2,\quad|R_N|\le\frac1{N+1}

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

Term count versus Partial sum. Partial sums, Limit ln2. Selected finite partial sum: 0.63452 . Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.000.520140Partial sumTerm count
Partial sumsLimit ln2 · dashed

finite partial sumS8=0.634524,∣R8∣≤19=0.111111S_8=0.634524,\quad|R_8|\le\frac1{9}=0.111111

Partial sums alternate around ln2. The next-term bound needs decreasing magnitudes tending to zero.

Assumptions, units, and limits

Term magnitudes decrease to zero. The harmonic absolute-value series diverges, so convergence is conditional.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. The alternating-series test requires magnitudes decreasing to zero. Absolute convergence is stronger and implies ordinary convergence. For this alternating harmonic series the next omitted magnitude bounds the remainder, while the absolute harmonic series diverges.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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