The nth Term Test for Divergence

Why convergent series have shrinking terms

A series a₁ + a₂ + a₃ + … converges when its partial sums Sₙ = a₁ + a₂ + … + aₙ approach a limit S. Each term is the gap between two partial sums next to each other: aₙ = Sₙ − Sₙ₋₁.

If the partial sums approach S, then for large n both Sₙ and Sₙ₋₁ are close to S, so their gap is close to S − S = 0. So the terms of a convergent series must tend to 0. Eventually the series is adding almost nothing.

The test

Turned around, that is the nth term test: if aₙ does not tend to 0, the series diverges.

Take Σ n/(2n + 1) = 1/3 + 2/5 + 3/7 + …. Divide the top and the bottom of the term by n: n/(2n + 1) = 1/(2 + 1/n). As n grows, 1/n tends to 0, so the term tends to 1/2, not 0. The series diverges. Adding nearly a half, again and again, passes every number: the partial sums are about 4.41 after 10 terms and 48.86 after 100.

The test also covers terms that have no limit at all. In 1 − 1 + 1 − 1 + …, the terms are 1 and −1 by turns, and the partial sums are 1, 0, 1, 0, …, which settle on nothing. The series diverges.

The terms n/(2n + 1) for n = 1 to 10, with the vertical scale stretched: 1/3, 2/5, 3/7, … They rise toward the gold line at 1/2, not down to 0.

The converse is false

Terms that tend to 0 do not make a series converge. The harmonic series 1 + 1/2 + 1/3 + 1/4 + … has terms 1/n, which tend to 0, and its partial sums still grow past every bound.

Group the terms in blocks that end at 2, 4, 8, 16, …: 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + …. In the block (1/3 + 1/4), each term is at least 1/4, so the block is at least 2 × 1/4 = 1/2. In the next block, each of the four terms is at least 1/8, so it is at least 4 × 1/8 = 1/2. Every block is at least 1/2.

So the sum of the first 2ᵏ terms is at least 1 + k/2. After 1024 terms, which is 2¹⁰, it is at least 1 + 5 = 6; the actual sum is about 7.51. Taking k large enough passes any number, so the harmonic series diverges, however slowly.

The harmonic partial sums at n = 1, 2, 4, 8, 16 and 32: about 1, 1.5, 2.08, 2.72, 3.38 and 4.06. The dashed curve passes through the guaranteed values 1, 1.5, 2, 2.5, 3 and 3.5 there: each doubling of n adds at least 1/2.

Only ever "diverges"

So the test has two outcomes, and only one is a verdict. If aₙ does not tend to 0, the series diverges. If aₙ does tend to 0, the test says nothing: the series might converge, like Σ 1/n², or diverge, like Σ 1/n. Both have terms tending to 0, and a different test must decide between them.

It is the first test to try, because it is the cheapest: one limit. When it says nothing, the shape of the term chooses the next test, such as the p-series, a comparison, or the alternating series test.

The usual mistakes

Reading terms that tend to 0 as proof of convergence. The harmonic series is the counterexample.

Taking the limit of the terms as the sum. The terms of Σ n/(2n + 1) tend to 1/2; the sum has no limit at all.

Saying the test proves Σ 1/n diverges. Its terms tend to 0, so the test is silent; the grouping argument shows it diverges.

Reading terms that tend to 0 as proof of divergence. Σ 1/n² converges.

A grant that settles

In the application below, a yearly grant tends to $5 thousand. The test rules the total out from settling, and adding the grants year by year finds when they pass the fund.

Worked example: An Endowment Whose Yearly Grant Settles but Whose Total Does Not: The Year the Fund Runs Dry

Question A trust pays a grant every year. The grant in year n is 5nn+1 thousand dollars. The trustee argues that the grants settle at $5 thousand a year, so the total paid out settles as well and an endowment of $100 thousand will last forever. (a) Use the nth term test to show that the total paid out does not settle. (b) In which year does the total paid out first pass $100 thousand?

1.Work out what one year's grant approaches. Divide the top and the bottom by n: 5nn+1 = 51 + 1n. As n grows, 1n tends to 0, so the grant tends to $5 thousand a year.

The grant in year n is 5nn+1 thousand dollars, and the bars are still rising.
2.That is what the nth term test asks about. If ∑ a_n converges then a_n must tend to 0. Here a_n tends to 5, so the condition fails and the series of grants cannot converge.

Dividing top and bottom by n gives 51 + 1n, which tends to 5.
3.(a) The total paid out diverges. The yearly grant settles at $5 thousand, but the running total of the grants grows without bound, and paying out nearly $5 thousand every year forever exhausts an endowment of any size.

(a) The terms tend to 5 and not to 0, so by the nth term test the total diverges.
4.For the year itself, add the grants one at a time. A rough guess is 100 ÷ 5 = 20 years, and the true answer is a little later because each grant is slightly under $5 thousand. The running totals are $86.8 thousand after 20 years, $96.3 thousand after 22 years and $101.1 thousand after 23 years.

The running total is $86.8 thousand after 20 years and $96.3 thousand after 22.
5.(b) The total paid out first passes $100 thousand in year 23. The endowment is exhausted then, not never, and the trustee's plan fails in its twenty-third year.

(b) In year 23 the total reaches $101.1 thousand and the endowment is gone.

Answer: (a) The grants tend to $5 thousand rather than to 0, so by the nth term test the total paid out diverges; (b) year 23

Common mistakes

Taking "the grants tend to $5 thousand" as a reason the total tends to $5 thousand. The limit of the terms and the sum of the series are two different quantities: here the terms settle at 5 while the sum grows without bound.
Reading the nth term test the other way round, as though terms tending to 0 would prove the total settles. The test can only rule a series out. The harmonic series ∑ 1n has terms tending to 0 and diverges all the same, so a limit of 0 decides nothing on its own.

More series and convergence problems, worked step by step →