core29 min studyLesson 172 in Mathematics

Comparison tests and positive series

Turn monotone partial sums into reliable comparison, condensation, and remainder estimates.

Before you begin

Limsup, liminf, and subsequences

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The question for today

What you’ll understand

Use bounded partial sums to characterize convergence of nonnegative series.
Choose direct comparison, limit comparison, or condensation from the term structure.
Derive a quantitative remainder bound rather than merely naming a test.

A tiny term does not guarantee a small accumulated total. Savings deposited in ever smaller amounts may still grow without bound, just as increasingly small numerical corrections may never have a finite sum. For nonnegative series, cancellation cannot obscure the picture: every partial sum increases. This makes convergence a question of whether the accumulating quantity stays bounded and gives comparison tests a transparent logical basis. The same reasoning turns a convergence proof into a usable approximation budget.

Positive terms produce monotone accounting

For nonnegative terms, partial sums are increasing, so completeness says they converge exactly when they have a finite upper bound. This is the foundation of direct comparison. If each late term of one series is at most a constant multiple of a summable nonnegative term, its tail sums are bounded. Finitely many earlier terms are irrelevant. Conversely, a lower bound by terms whose partial sums grow without bound proves divergence.

The condition that terms approach zero is only necessary. In the harmonic series, group terms from just after one power of two through the next power. Every such block contributes at least one half, so infinitely many blocks force unlimited accumulation. This argument identifies the mechanism of divergence rather than relying on an imprecise claim that reciprocals shrink slowly. It also suggests why grouping by geometric scales is often more informative than inspecting individual terms.

0\le a_n\le Cb_n\ (n\ge N),\quad\sum_{n=N}^{\infty}b_n<\infty\quad\Longrightarrow\quad\sum_{n=N}^{\infty}a_n<\infty

The comparison needs eventual nonnegativity and a fixed finite C; it does not require an inequality for the finite initial segment.↔ Scroll wide equations horizontally.

Match the dominant scale before choosing a test

Limit comparison formalizes the idea that terms share a scale. If the ratio of two positive term sequences approaches a finite positive number, each eventually bounds the other up to constants, so their series converge or diverge together. For a rational expression in n, dividing numerator and denominator by their highest powers often reveals the appropriate p-series. A zero ratio gives only a one-sided comparison; an infinite ratio also needs the correct direction.

Logarithmic corrections require finer resolution. For decreasing nonnegative terms, Cauchy condensation groups dyadic blocks and compares the original series with the sum of powers of two times the terms at those powers. Applied to reciprocal n times a power of log n, condensation produces a constant multiple of a p-series in the block index. The resulting threshold is sensitive to the logarithmic exponent even though every such original term resembles a harmonic term on a coarse plot.

\sum_{n=2}^{\infty}\frac{1}{n(\log n)^p}<\infty\quad\Longleftrightarrow\quad p>1

For real p, condensation applies after the terms become decreasing; log denotes the natural logarithm. The condensed terms are constant multiples of k^(−p).↔ Scroll wide equations horizontally.

Convergence should give an error estimate

For a positive decreasing integrable function matching the terms at integers, rectangles bound a series tail between two improper integrals. This proves the p-series threshold and estimates how many terms meet a tolerance. The distinction between convergence and rapid convergence matters: a series may converge while demanding impractically many terms for a tight error budget. An exact answer to the convergence question alone says little about computational efficiency.

In the spectral lab, the coefficient of mode n has magnitude n to the negative p. Since sine never exceeds one in absolute value, a convergent coefficient tail supplies a uniform error bound for omitted modes at every position and time. Predict how increasing p changes this bound before adjusting the control. The display shows only finitely many modes; the comparison with an infinite positive series supplies the guarantee that the finite picture itself cannot provide.

Work it through

Certifying a rational-series approximation

Prove convergence of $\sum_{n=1}^{\infty}1/(n^2+3n)$ and bound the error after N terms.

Every term is positive and satisfies $1/(n^2+3n)\le1/n^2$ . The p-series with exponent 2 converges, so direct comparison proves convergence.
For the remainder beyond N, compare with $\sum_{n=N+1}^{\infty}1/n^2$ . Since x↦1/x² decreases, each term is bounded above by the integral over the preceding unit interval.
Adding these inequalities gives a remainder at most $\int_N^{\infty}x^{-2}\,dx=1/N$ . Thus N=200 guarantees error at most 0.005 without knowing the exact infinite sum.

The result

The series converges, and truncating after N terms loses no more than 1/N. A bound can be useful even when it is not sharp.

The living notebook

Explore the model

Interactive lab

Interactive lab · spectral

Bound omitted coefficients instead of trusting resolution

The controls set N and p in Σ sin(nx−t)/nᵖ. Predict whether the absolute coefficient tail is summable for p>1 and compare profiles as N grows. The lab displays finite sums, not the infinite tail or a proof of convergence.

Finite sumHighest included mode

Sum at x = π/2: 0.867
Smallest wavelength: 2π / 5
Highest-mode coefficient: 0.2

Mode count N5Coefficient decay p1

t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA scalar positive-series bound can certify a function approximation uniformly.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The ratio gives an upper comparison with a divergent series, which is inconclusive. Taking bₙ=1/n, both aₙ=1/n² and aₙ=1/(n log n) have ratio zero but different convergence behavior.

Your turn

Try it yourself

Determine convergence of $\sum_{n=2}^{\infty}1/(n(\log n)^2)$ and explain why comparison only with 1/n is insufficient.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Group indices from 2ᵏ through 2ᵏ⁺¹−1 or apply condensation.

Compare with the solution

The terms are positive and decreasing. Condensation replaces them by $2^k/[2^k(\log 2^k)^2]=1/[k^2(\log2)^2]$ , a convergent constant multiple of a p-series. Therefore the original series converges. Its terms being below 1/n would not prove this, because an upper bound by a divergent positive series supplies no finite bound on the original partial sums.

Keep the idea

Take it with you

Nonnegative series converge exactly when their partial sums are bounded.
Comparison direction matters when a benchmark diverges or a ratio vanishes.
Remainder estimates connect infinite convergence to finite computation.

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