Math Infinitum

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Intuition

When no term is negative the partial sums can only rise. A rising sequence has just two possible fates: it passes every bound, or it settles. That reduces a question about a series to a question about whether one number can be found above all the running totals, which is often far easier to answer.

Water poured into a tall glass, never removed. Either it overflows or it reaches a level. Nothing else can happen, and knowing the glass has a rim is enough to know a level is reached.

With no negative term the running totals can only rise, which leaves exactly two fates. The totals of $\sum 1/k^2$ stay under the rim and therefore settle; those of $\sum 1/k$ climb past that rim and past every other one you draw. Bounded above is the whole of the question.

The bounded partial sums criterion

This is the monotone convergence theorem read for series. It is the workhorse of the chapter, and every later test reduces to it by comparison.

\begin{aligned}&a_k \ge 0 \implies \big( \sum a_k \text{ converges}\\ &\quad \iff (s_n) \text{ is bounded above} \big)\end{aligned}

What it gives

For non-negative terms the partial sums are increasing, so only two fates remain: they pass every bound, or they settle.
So $\sum a_k$ with $a_k\ge0$ converges exactly when its partial sums are bounded above, and then the sum is their supremum.

Non-negative series converge exactly when bounded

\begin{aligned}&a_k \ge 0 \implies \big( \sum a_k \text{ converges}\\ &\quad \iff (s_n) \text{ bounded above} \big)\end{aligned}

Non-negative terms make the partial sums increasing, and an increasing sequence converges precisely when it is bounded above. Each direction is one theorem from the previous chapter.

Proof steps

s_{n+1} - s_n = a_{n+1} \ge 0, \text{ so } (s_n) \text{ is increasing}

Each new term can only add, never take away.

\text{If } (s_n) \text{ is bounded above it converges}

This is the monotone convergence theorem, applied to the partial sums.

\text{If } (s_n) \text{ converges it is bounded}

Every convergent sequence is bounded, which was proved in the previous chapter.

\text{So convergence and boundedness agree}

Each implies the other, which is what the statement claims.

Applications

Practice

Only two fates

a_k \ge 0 \implies s_{n+1} \ge s_n

Each new term adds something non-negative, so the totals cannot go back down.

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A series has non-negative terms and bounded partial sums. What follows?

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The sign condition is not decorative

$\sum (-1)^k$ has bounded partial sums and does not converge.

The totals alternate between two values, staying bounded without ever settling.

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A series has bounded partial sums but terms of both signs. Does it converge?

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Which theorem from the previous chapter does this criterion rest on?

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Blocks That Never Shrink

Group the harmonic series into blocks of doubling length. Each block has more terms as it goes, and they shrink at exactly the rate that keeps each total the same.

\tfrac13+\tfrac14>\tfrac14+\tfrac14=\tfrac12

\tfrac15+\tfrac16+\tfrac17+\tfrac18>4\cdot\tfrac18=\tfrac12

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Why does grouping show that $\sum\tfrac1k$ diverges?

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Grouping the harmonic series shows $s_{2^m}\ge1+\tfrac m2$ . What is the smallest $m$ for which that bound guarantees $s_{2^m}>3$ ?

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Brackets may be inserted freely into any convergent series without changing its sum.

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A series of non-negative terms has partial sums $0.9,\ 0.99,\ 0.999,\dots$ , each with one more nine than the one before. What is its sum?

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A series of non-negative terms either converges or has partial sums passing every bound.

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What You Learned

Non-negative terms make the partial sums increase, leaving only two fates.
Such a series converges exactly when its partial sums are bounded above.
The criterion rests on monotone convergence and so on completeness.
Grouping the harmonic series into doubling blocks shows it diverges.

Final checkpoint

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A series of non-negative terms has partial sums that never exceed $7$ . What follows?

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Completion

Lesson complete

Great work! You now know how to:

State the bounded partial sums criterion and name the theorem behind it
Prove that the harmonic series diverges by grouping
Say why non-negative terms are needed for any of this