Real Analysis 18 | Leibniz Criterion [dark version]

Q: Why does the harmonic series fail even though its terms go to zero?

For $\sum_{k=1}^\infty 1/k$, the terms satisfy $1/k\to 0$, but the partial sums still grow without bound. The rectangle visualization shows each added rectangle has positive area, so the total area diverges to infinity. Leibniz Criterion fixes this by introducing alternating signs, which create cancellation rather than one-direction growth.

Q: What conditions on $a_k$ make Leibniz Criterion work for $\sum (-1)^{k+1} a_k$?

The test requires $a_k\ge 0$, $a_k$ monotonically decreasing, and $\lim_{k\to\infty} a_k=0$. With these, the alternating series has partial sums that oscillate but with shrinking amplitude.

Q: How do even and odd partial sums behave in the proof?

Let $S_n$ be the partial sums. The even-indexed subsequence $S_{2L}$ is monotone decreasing, while the odd-indexed subsequence $S_{2L+1}$ is monotone increasing. This comes from the fact that $a_k$ decreases: adding the next term changes the partial sum by a smaller magnitude than before.

Q: How does the proof show the two subsequences squeeze each other?

Comparing consecutive subsequences gives $S_{2L+1}-S_{2L} = -a_{2L+1}\le 0$. Equivalently, $S_{2L}\le S_{2L+1}$ (with the sign convention used in the derivation). This establishes bounds between the increasing and decreasing subsequences, so both are trapped in an interval.

Q: How is Leibniz Criterion applied to $\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k}}$?

Write it as $\sum (-1)^{k+1} a_k$ with $a_k=1/\sqrt{k}$. The sequence $1/\sqrt{k}$ is nonnegative, decreases monotonically, and tends to zero. Therefore the alternating series converges to some finite real number.

TL;DR

Leibniz Criterion is a sufficient convergence test for alternating series $\sum (- 1)^{k + 1} a_{k}$ .

Briefing Cornell Notes

Briefing

Leibniz Criterion (also called the alternating series test or Leibniz’s test) supplies the missing sufficient condition for when an infinite series with alternating signs actually converges. The key idea is that alternating terms can “cancel” enough to overcome divergence that would occur if all terms were added with the same sign—something the harmonic series illustrates starkly: the partial sums of $\sum_{k = 1}^{\infty} 1/ k$ grow without bound even though $1/ k \to 0$ . In other words, having terms that go to zero is necessary for convergence but not sufficient.

The criterion’s visualization flips the harmonic picture by alternating rectangle areas: positive contributions and negative contributions alternate, so the running total no longer drifts monotonically upward. Formally, start with a sequence $a_{k} \geq 0$ that decreases monotonically to zero. Consider the alternating series $\sum_{k = 1}^{\infty} (- 1)^{k + 1} a_{k}$ . Leibniz Criterion guarantees that the sequence of partial sums converges to a finite real limit.

The proof strategy turns the partial sums into two subsequences—one capturing the even indices and one capturing the odd indices. Because $a_{k}$ is nonnegative and monotonically decreasing, the even-indexed partial sums form a monotone decreasing sequence, while the odd-indexed partial sums form a monotone increasing sequence. Moreover, the two subsequences squeeze each other: comparing $S_{2 L}$ and $S_{2 L + 1}$ shows their difference is exactly $- a_{2 L + 1}$ , which is $\leq 0$ . This yields inequalities that bound both subsequences.

Once both subsequences are monotone and bounded, each must converge by the monotone convergence criterion for sequences. The remaining step uses the fact that the gap between the two subsequences is $a_{2 L + 1}$ , which tends to zero as $L \to \infty$ . That forces the two subsequential limits to coincide, so the entire sequence of partial sums $S_{n}$ converges to a single real number $S$ . The result matches the intuition from the rectangle picture: alternating signs create a controlled oscillation whose amplitude shrinks to zero.

A concrete application closes the lesson: the series $\sum_{k = 1}^{\infty} \frac{( - 1 ) ^{k}}{k}$ fits the criterion because $a_{k} = 1/ k$ decreases to zero. Therefore the series converges (even though its sum is not computed directly). The practical takeaway is straightforward: when a series alternates and the magnitudes decrease monotonically to zero, Leibniz Criterion provides convergence without needing to evaluate the limit explicitly.

Cornell Notes

Leibniz Criterion gives a convergence test for alternating series $\sum (- 1)^{k + 1} a_{k}$ . If $a_{k}$ is nonnegative, monotonically decreasing, and $a_{k} \to 0$ , then the partial sums $S_{n}$ converge to a finite real number. The proof splits $S_{n}$ into even and odd subsequences: even partial sums decrease, odd partial sums increase, and the two subsequences are bounded and squeeze together. The shrinking gap between them equals $a_{2 L + 1}$ , which goes to zero, forcing both subsequences to share the same limit. This makes the test a reliable way to confirm convergence without computing the sum.

Why does the harmonic series fail even though its terms go to zero?

For

\sum_{k = 1}^{\infty} 1/ k

, the terms satisfy

1/ k \to 0

, but the partial sums still grow without bound. The rectangle visualization shows each added rectangle has positive area, so the total area diverges to infinity. Leibniz Criterion fixes this by introducing alternating signs, which create cancellation rather than one-direction growth.

What conditions on $a_{k}$ make Leibniz Criterion work for $\sum (- 1)^{k + 1} a_{k}$ ?

The test requires

a_{k} \geq 0

a_{k}

monotonically decreasing, and

lim_{k \to \infty} a_{k} = 0

. With these, the alternating series has partial sums that oscillate but with shrinking amplitude.

How do even and odd partial sums behave in the proof?

Let

S_{n}

be the partial sums. The even-indexed subsequence

S_{2 L}

is monotone decreasing, while the odd-indexed subsequence

S_{2 L + 1}

is monotone increasing. This comes from the fact that

a_{k}

decreases: adding the next term changes the partial sum by a smaller magnitude than before.

How does the proof show the two subsequences squeeze each other?

Comparing consecutive subsequences gives

S_{2 L + 1} - S_{2 L} = - a_{2 L + 1} \leq 0

. Equivalently,

S_{2 L} \leq S_{2 L + 1}

(with the sign convention used in the derivation). This establishes bounds between the increasing and decreasing subsequences, so both are trapped in an interval.

Why does convergence of $a_{k} \to 0$ force the whole series to converge?

Both subsequences (even and odd partial sums) converge because they are monotone and bounded. Their difference equals

a_{2 L + 1}

, and since

a_{2 L + 1} \to 0

, the two subsequential limits must match. That makes the full sequence

S_{n}

converge to a single limit

S

How is Leibniz Criterion applied to $\sum_{k = 1}^{\infty} \frac{( - 1 ) ^{k}}{k}$ ?

Write it as

\sum (- 1)^{k + 1} a_{k}

with

a_{k} = 1/ k

. The sequence

1/ k

is nonnegative, decreases monotonically, and tends to zero. Therefore the alternating series converges to some finite real number.

Review Questions

State Leibniz Criterion precisely: what three properties must $a_{k}$ satisfy for an alternating series to converge?
In the proof, why are the even-indexed partial sums monotone decreasing and the odd-indexed partial sums monotone increasing?
For an alternating series $\sum (- 1)^{k + 1} a_{k}$ , what quantity measures the gap between the even and odd subsequential limits, and why does it go to zero?

Key Points

1
Leibniz Criterion is a sufficient convergence test for alternating series $\sum (- 1)^{k + 1} a_{k}$ .
2
If $a_{k} \geq 0$ , $a_{k}$ is monotonically decreasing, and $a_{k} \to 0$ , then the partial sums $S_{n}$ converge.
3
Alternating signs can create cancellation strong enough to turn a would-be divergent “positive-only” behavior into convergence.
4
The proof splits partial sums into even and odd subsequences, which become monotone (one decreasing, one increasing).
5
Both subsequences are bounded, so each converges by the monotone convergence criterion for sequences.
6
The difference between the even and odd subsequences equals $a_{2 L + 1}$ , which tends to zero, forcing a common limit.
7
A series like $\sum_{k = 1}^{\infty} \frac{( - 1 ) ^{k}}{k}$ converges because $1/ k$ decreases to zero.

Highlights

Leibniz Criterion turns the “terms go to zero” necessity into a practical sufficient condition for alternating series.

Even partial sums decrease while odd partial sums increase, and the two sequences squeeze together.

The gap between the subsequences is exactly a2L+1​, so convergence of ak​→0 completes the argument.

The harmonic series diverges because all terms are positive, but alternating signs can prevent runaway growth.

The test confirms convergence for ∑k=1∞​k​(−1)k​ without computing its sum.

Topics

Leibniz Criterion
Alternating Series Test
Monotone Convergence
Partial Sums
Harmonic Series