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Convergence tests

Infinity and strange math · level 8, lesson 2 of 7

A question to start

Two infinite sums look almost the same:

\[1 + \frac12 + \frac13 + \frac14 + \cdots \qquad\text{and}\qquad 1 + \frac14 + \frac19 + \frac1{16} + \cdots\]

In level 4 you proved that the first one, the harmonic series, grows without limit. The second adds the reciprocals of the square numbers. Its terms shrink faster. Does it have a finite total?

It does, and in the 1730s Leonhard Euler astonished Europe by finding it exactly:

\[1 + \frac14 + \frac19 + \frac1{16} + \cdots = \frac{\pi^2}{6} \approx 1.6449.\]

A circle number from a sum of squares! Euler's proof is beyond this lesson, but proving that the sum is finite is not. This lesson gives you the first tools for deciding whether an infinite sum adds up to a number: the convergence tests.

The idea

Series, partial sums and convergence

An infinite sum \(a_1 + a_2 + a_3 + \cdots\) is called a series, written

\[\sum_{n=1}^{\infty} a_n.\]

Nobody can do infinitely many additions, so we define the sum through a limit. The partial sums are

\[S_1 = a_1, \quad S_2 = a_1 + a_2, \quad S_3 = a_1 + a_2 + a_3, \quad \ldots, \quad S_N = \sum_{n=1}^{N} a_n.\]

The series converges to \(S\) if \(\lim_{N \to \infty} S_N = S\), in the exact \(\varepsilon\)-\(N\) sense of the last lesson. Otherwise it diverges. For example the geometric series \(\tfrac12 + \tfrac14 + \tfrac18 + \cdots\) has \(S_N = 1 - \tfrac1{2^N}\), which has limit 1, so the series converges to 1.

Test 0: the terms must shrink to zero

If \(\sum a_n\) converges to \(S\), then \(a_N = S_N - S_{N-1}\), and both \(S_N\) and \(S_{N-1}\) approach \(S\), so \(a_N\) approaches \(S - S = 0\). So:

Divergence test. If the terms \(a_n\) do not tend to 0, the series diverges.

For example \(\sum \tfrac{n}{n+1} = \tfrac12 + \tfrac23 + \tfrac34 + \cdots\) diverges, because its terms tend to 1.

But the test only works one way. The harmonic series has terms tending to 0 and still diverges. Terms shrinking to 0 is necessary for convergence, never sufficient.

Why positive series are easier

If every term is positive, the partial sums only go up: \(S_1 < S_2 < S_3 < \cdots\) Such a sequence can do only one of two things:

  • grow without limit (diverge to infinity), or
  • stay below some ceiling, and then it converges.

The second point is a basic property of the real numbers, called completeness (the monotone convergence theorem): an increasing sequence with a ceiling has a limit. It is exactly the property the rationals lack (the increasing decimals \(1, 1.4, 1.41, 1.414, \ldots\) have no rational limit, since \(\sqrt2\) is irrational). So for a positive series, convergence is just a question of whether the partial sums stay bounded.

Test 1: the comparison test

Comparison test. Suppose \(0 \le a_n \le b_n\) for every \(n\).

  • If \(\sum b_n\) converges, then \(\sum a_n\) converges too.
  • If \(\sum a_n\) diverges, then \(\sum b_n\) diverges too.

Why. Every partial sum of \(\sum a_n\) is at most the matching partial sum of \(\sum b_n\), which is at most the total \(B\) of \(\sum b_n\). So the partial sums of \(\sum a_n\) increase and stay below \(B\), and by completeness they converge. The second part is the same fact turned round: if the smaller one grows without limit, so does the bigger.

\(\sum 1/n^2\) converges

Compare \(\tfrac1{n^2}\) with something we can add exactly. For \(n \ge 2\),

\[\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}.\]

(Check the last step: \(\tfrac1{n-1} - \tfrac1n = \tfrac{n - (n-1)}{n(n-1)} = \tfrac1{n(n-1)}\).) Adding these from \(n = 2\) to \(N\), almost everything cancels, like a folding telescope:

\[\left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \cdots + \left(\tfrac1{N-1} - \tfrac1N\right) = 1 - \frac1N < 1.\]

So \(S_N = 1 + \sum_{n=2}^{N} \tfrac1{n^2} < 1 + 1 = 2\) for every \(N\). The partial sums increase and stay below 2, so the series converges, to some number at most 2. Euler's value \(\tfrac{\pi^2}6 \approx 1.645\) is indeed below 2.

Test 2: the ratio test

The geometric series \(\sum r^n\) converges when \(0 \le r < 1\) and diverges when \(r \ge 1\). The ratio test says that a series whose terms eventually behave like a geometric series does the same.

Ratio test. For a series of positive terms, let \(r = \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n}\).

  • If \(r < 1\), the series converges.
  • If \(r > 1\), the series diverges.
  • If \(r = 1\), the test tells you nothing.

Why (for \(r < 1\)). Pick a number \(s\) with \(r < s < 1\). Since the ratios approach \(r\), from some point \(N\) on each ratio is less than \(s\), so \(a_{N+1} < s\,a_N\), \(a_{N+2} < s^2 a_N\), and so on. From then on the terms are smaller than those of the geometric series \(a_N(s + s^2 + s^3 + \cdots)\), which converges because \(s < 1\). By comparison, our series converges. (For \(r > 1\) the terms eventually grow, so they cannot tend to 0.)

The \(r = 1\) case really is undecided: for \(\sum \tfrac1n\) (diverges) and \(\sum \tfrac1{n^2}\) (converges) the ratios \(\tfrac{n}{n+1}\) and \(\tfrac{n^2}{(n+1)^2}\) both tend to 1.

Alternating series

Now let the signs take turns. The alternating harmonic series is

\[1 - \frac12 + \frac13 - \frac14 + \frac15 - \cdots\]

Its partial sums jump right and left, each jump smaller than the last:

ln 2 ≈ 0.693 0.5 1 S1 S2 S3 S4 S5 S6 S7 S8

Alternating series test (Leibniz). If the terms alternate in sign, shrink in size, and tend to 0, the series converges.

Why. The odd partial sums \(S_1, S_3, S_5, \ldots\) go down (each step subtracts a bigger amount than it then adds back), the even ones \(S_2, S_4, S_6, \ldots\) go up, and every even one is below every odd one. So the two families close in on each other from both sides, and the gap between neighbours, the size of a term, tends to 0. They squeeze in on one number. As a bonus, the true sum always lies between any two consecutive partial sums, so the error after \(N\) terms is less than the next term.

For the alternating harmonic series the limit is \(\ln 2 \approx 0.6931\), where \(\ln\) is the natural logarithm. (The proof uses the series for \(\ln(1 + x)\), from university calculus.)

A warning from the edge

The alternating harmonic series converges only because of the cancelling: without the minus signs it is the harmonic series, which diverges. Such a series is called conditionally convergent, and it behaves strangely. Rearrange the same terms as "one positive, then two negatives":

\[1 - \frac12 - \frac14 + \frac13 - \frac16 - \frac18 + \frac15 - \frac1{10} - \frac1{12} + \cdots\]

Every term is used exactly once, yet the sum is now \(\tfrac12 \ln 2\), half as much. (Group in threes: \(\left(1 - \tfrac12\right) - \tfrac14 = \tfrac12 - \tfrac14\), \(\left(\tfrac13 - \tfrac16\right) - \tfrac18 = \tfrac16 - \tfrac18\), and so on, which is half of the original series.) Bernhard Riemann proved that such a series can be rearranged to add up to any number you like. For infinite sums, the order can matter.

Worked examples

1. The first partial sums of \(\sum 1/n^2\). \(S_1 = 1\), \(S_2 = 1.25\), \(S_3 \approx 1.3611\), \(S_4 \approx 1.4236\), \(S_5 \approx 1.4636\). They climb slowly towards \(1.6449\ldots\), always below 2.

2. Comparison. Does \(\sum \dfrac{1}{2^n + 1}\) converge? Each term is positive and \(\tfrac{1}{2^n + 1} < \tfrac{1}{2^n}\). The series \(\sum \tfrac1{2^n} = \tfrac12 + \tfrac14 + \cdots\) converges (to 1). By comparison, so does ours, to something less than 1.

3. Comparison the other way. Does \(\sum \dfrac{1}{\sqrt n}\) converge? For \(n \ge 1\), \(\sqrt n \le n\), so \(\tfrac{1}{\sqrt n} \ge \tfrac1n\). The harmonic series diverges, so the bigger series \(\sum \tfrac{1}{\sqrt n}\) diverges too.

4. Ratio test. Does \(\sum \dfrac{n}{2^n} = \tfrac12 + \tfrac24 + \tfrac38 + \cdots\) converge? \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)/2^{n+1}}{n/2^n} = \dfrac{n+1}{2n}\), which tends to \(\tfrac12 < 1\). It converges. (Its sum is in fact 2.)

5. Ratio test with factorials. Does \(\sum \dfrac{3^n}{n!}\) converge? \(\dfrac{a_{n+1}}{a_n} = \dfrac{3^{n+1}}{(n+1)!} \cdot \dfrac{n!}{3^n} = \dfrac{3}{n+1}\), which tends to \(0 < 1\). It converges. (Starting from \(n = 1\) its sum is \(e^3 - 1 \approx 19.0855\).) Factorials beat powers.

6. Bracketing \(\ln 2\). The partial sums of the alternating harmonic series are \(S_1 = 1\), \(S_2 = 0.5\), \(S_3 \approx 0.8333\), \(S_4 \approx 0.5833\), \(S_5 \approx 0.7833\), \(S_6 \approx 0.6167\). The true sum lies between \(S_5\) and \(S_6\), and \(\ln 2 \approx 0.6931\) does.

Try it

  1. Use the divergence test to show that \(\sum \dfrac{2n}{n + 5}\) diverges.
  2. Use comparison to show that \(\sum \dfrac{1}{n^2 + 1}\) converges.
  3. Use comparison to show that \(\sum \dfrac{1}{2n - 1} = 1 + \tfrac13 + \tfrac15 + \cdots\) diverges.
  4. Use the ratio test on \(\sum \dfrac{n^2}{3^n}\).
  5. Use the ratio test on \(\sum \dfrac{n!}{10^n}\).
  6. What does the ratio test say about \(\sum \dfrac{1}{n^3}\)? Does the series converge?
  7. How many terms of the alternating harmonic series guarantee an error less than \(0.01\)?
  8. Challenge: Leibniz's series \(1 - \tfrac13 + \tfrac15 - \tfrac17 + \cdots\) is known to equal \(\tfrac\pi4\). Explain why it converges, and use the first four partial sums to bracket \(\pi\).
Answers
  1. \(\tfrac{2n}{n + 5} = \tfrac{2}{1 + 5/n}\), which tends to 2, not 0. So the series diverges.
  2. \(0 < \tfrac1{n^2 + 1} < \tfrac1{n^2}\), and \(\sum \tfrac1{n^2}\) converges, so by comparison so does \(\sum \tfrac1{n^2 + 1}\).
  3. \(2n - 1 < 2n\), so \(\tfrac1{2n - 1} > \tfrac1{2n}\). And \(\sum \tfrac1{2n} = \tfrac12 \sum \tfrac1n\) diverges (half of something unbounded is unbounded). So \(\sum \tfrac1{2n-1}\) diverges.
  4. \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^2}{3^{n+1}} \cdot \dfrac{3^n}{n^2} = \dfrac13 \left(\dfrac{n+1}{n}\right)^2\), which tends to \(\tfrac13 < 1\). It converges. (Its sum is \(1.5\).)
  5. \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{10^{n+1}} \cdot \dfrac{10^n}{n!} = \dfrac{n+1}{10}\), which tends to infinity, more than 1. It diverges (the terms grow once \(n \ge 10\)).
  6. \(\dfrac{n^3}{(n+1)^3}\) tends to 1, so the ratio test says nothing. But \(\tfrac1{n^3} \le \tfrac1{n^2}\) and \(\sum \tfrac1{n^2}\) converges, so by comparison \(\sum \tfrac1{n^3}\) converges.
  7. The error after \(N\) terms is less than the next term, \(\tfrac1{N+1}\). We need \(\tfrac1{N+1} \le 0.01\), so \(N + 1 \ge 100\), \(N = 99\) terms.
  8. The terms alternate, shrink (\(1 > \tfrac13 > \tfrac15 > \cdots\)) and tend to 0, so the alternating series test applies. Partial sums: \(S_1 = 1\), \(S_2 = \tfrac23 \approx 0.6667\), \(S_3 = \tfrac{13}{15} \approx 0.8667\), \(S_4 = \tfrac{76}{105} \approx 0.7238\). The sum \(\tfrac\pi4\) is between \(S_4\) and \(S_3\), so \(\pi\) is between \(4 \times 0.7238 \approx 2.895\) and \(4 \times 0.8667 \approx 3.467\). True, but slow: this series is a terrible way to compute \(\pi\).

Common mistake

"The terms go to 0, so it converges." The harmonic series is the standing counterexample. Terms tending to 0 is needed but not enough.

Using the ratio test when the limit is 1. A limit of exactly 1 tells you nothing; switch to comparison (or another test). Also, the ratio test needs the limit of the ratios: "each ratio is less than 1" is not enough (\(\tfrac{n}{n+1} < 1\) for the divergent harmonic series).

Where it leads