Unit 7, lesson 100

Absolute and Conditional Convergence

The difference between series that converge because their terms are small and series that converge only through cancellation, and why only the first kind can be reordered safely.

  • Theory part 199
  • Practice part 200
  • Convergence and Convergence Tests

For a finite sum, the order of the numbers never matters. For an infinite series it can. Some series converge so solidly that you can shuffle their terms any way you like and still get the same total. Others converge only because positive and negative terms cancel in a delicate balance, and reordering them can change the sum to anything you want. So before you reorder the terms of a series, you need to know which kind it is.

Key terms

Absolute convergence

Definition A series ∑an\sum a_n converges absolutely if the series of absolute values

∑n=1∞∣an∣\sum_{n=1}^{\infty} \lvert a_n \rvert

converges. For a series with nonnegative terms, absolute convergence is the same as convergence.

In plain words Throw away all the minus signs. If the series still converges, it converges absolutely.

Example The series ∑(−1)nn2\sum \frac{(-1)^n}{n^2} converges absolutely, because ∑∣(−1)nn2∣=∑1n2\sum \left\lvert \frac{(-1)^n}{n^2} \right\rvert = \sum \frac{1}{n^2} is a convergent p-series.

Absolute convergence implies convergence

Definition If ∑∣an∣\sum \lvert a_n \rvert converges, then ∑an\sum a_n converges, and

∣∑n=1∞an∣≤∑n=1∞∣an∣.\left\lvert \sum_{n=1}^{\infty} a_n \right\rvert \le \sum_{n=1}^{\infty} \lvert a_n \rvert.

The proof uses 0≤an+∣an∣≤2∣an∣0 \le a_n + \lvert a_n \rvert \le 2\lvert a_n \rvert. By direct comparison ∑(an+∣an∣)\sum (a_n + \lvert a_n \rvert) converges, and then

∑an=∑(an+∣an∣)−∑∣an∣\sum a_n = \sum (a_n + \lvert a_n \rvert) - \sum \lvert a_n \rvert

converges by linearity, as the difference of two convergent series.

In plain words If the series converges even without any help from cancellation, putting the signs back cannot spoil that.

Example For ∑sin⁡nn2\sum \frac{\sin n}{n^2} the signs follow no simple pattern, so the alternating series test does not apply. But ∣sin⁡nn2∣≤1n2\left\lvert \frac{\sin n}{n^2} \right\rvert \le \frac{1}{n^2}, so the series converges absolutely by comparison, and therefore it converges.

Conditional convergence

Definition A series ∑an\sum a_n converges conditionally if ∑an\sum a_n converges but ∑∣an∣\sum \lvert a_n \rvert diverges.

Every series falls into exactly one of three classes. It converges absolutely, it converges conditionally, or it diverges.

In plain words A conditionally convergent series only converges because its positive and negative terms cancel. Without the signs it would blow up.

Example The alternating harmonic series ∑(−1)n−1n\sum \frac{(-1)^{n-1}}{n} converges by the alternating series test, but ∑1n\sum \frac{1}{n} diverges. So it converges conditionally.

Rearrangements

Definition A rearrangement of ∑an\sum a_n is a series ∑k=1∞aσ(k)\sum_{k=1}^{\infty} a_{\sigma(k)}, where σ\sigma is a bijection of the positive integers onto themselves. Every term ana_n appears exactly once in the new series, in a new position.

In plain words Use exactly the same terms, each one once, just in a different order.

Example The alternating harmonic series written in the order of one positive term followed by two negative terms is the rearrangement

1−12−14+13−16−18+15−110−112+⋯ .1 - \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{6} - \frac{1}{8} + \frac{1}{5} - \frac{1}{10} - \frac{1}{12} + \cdots.

Each odd reciprocal 12k−1\frac{1}{2k - 1} is followed by −14k−2-\frac{1}{4k - 2} and −14k-\frac{1}{4k}, so every term of the original series appears exactly once.

Absolute convergence survives rearrangement

Definition If ∑an\sum a_n converges absolutely with sum SS, then every rearrangement ∑aσ(k)\sum a_{\sigma(k)} also converges absolutely, and its sum is SS.

In plain words Absolutely convergent series behave like finite sums. You can reorder them, group the positive terms and the negative terms, and the total stays the same.

Example The geometric series

∑n=0∞(−12)n=1−12+14−18+⋯=11+1/2=23\sum_{n=0}^{\infty} \left(-\frac{1}{2}\right)^n = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots = \frac{1}{1 + 1/2} = \frac{2}{3}

converges absolutely. Adding all the positive terms first and then all the negative terms gives

(1+14+116+⋯ )−(12+18+⋯ )=43−23=23,\left(1 + \frac{1}{4} + \frac{1}{16} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{8} + \cdots\right) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3},

the same total.

Rearranging a conditionally convergent series

Definition If ∑an\sum a_n converges conditionally, then for every real number LL there is a rearrangement that converges to LL, and there are rearrangements whose partial sums tend to ∞\infty or to −∞-\infty. This is the Riemann rearrangement theorem. It works because the positive terms alone and the negative terms alone both form divergent series, while the terms themselves tend to 00.

In plain words With a conditionally convergent series, the order is part of the answer. Change the order and you can steer the sum anywhere.

Example Group the rearrangement from above in threes. The kkth group is

12k−1−14k−2−14k=14k−2−14k=12(12k−1−12k).\frac{1}{2k - 1} - \frac{1}{4k - 2} - \frac{1}{4k} = \frac{1}{4k - 2} - \frac{1}{4k} = \frac{1}{2}\left(\frac{1}{2k - 1} - \frac{1}{2k}\right).

So the sum of the first KK groups is half the partial sum s2Ks_{2K} of the alternating harmonic series, which tends to 12ln⁡2\frac{1}{2}\ln 2. Partial sums that end inside a group differ from these by at most two terms, which tend to 00. The rearranged series converges to 12ln⁡2≈0.3466\frac{1}{2}\ln 2 \approx 0.3466, half of the original sum ln⁡2\ln 2.

Check yourself

  1. Classify ∑n=1∞(−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}.
Show the answer

It converges by the alternating series test, but ∑1n\sum \frac{1}{\sqrt{n}} is a divergent p-series. So it converges conditionally.

  1. Classify ∑n=1∞(−1)nn2n\sum_{n=1}^{\infty} \frac{(-1)^n n}{2^n}.
Show the answer

The absolute values satisfy

n2n≤1n2for n≥10,\frac{n}{2^n} \le \frac{1}{n^2} \quad \text{for } n \ge 10,

since n3≤2nn^3 \le 2^n from there on. It holds at n=10n = 10, where 1000≤10241000 \le 1024, and each step multiplies n3n^3 by (n+1n)3≤1.331\left(\frac{n + 1}{n}\right)^3 \le 1.331 but 2n2^n by 22. So ∑n2n\sum \frac{n}{2^n} converges and the series converges absolutely.

  1. Classify ∑n=1∞cos⁡(nπ)nn+1\sum_{n=1}^{\infty} \cos(n\pi) \frac{n}{n + 1}.
Show the answer

Since cos⁡(nπ)=(−1)n\cos(n\pi) = (-1)^n, the terms have size nn+1→1\frac{n}{n + 1} \to 1 and do not tend to 00. The series diverges.

How this lesson connects