Calculus 2

The nth term test for divergence

A series whose terms stay near some fixed size, say three quarters, keeps piling up and can never settle on a total. What exactly must the terms of a series do for it to have a sum, and how much does that one fact decide?

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Infinite series and partial sumsCalculus 2

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The nth term test for divergence

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Why this exists

Iroda models the growth of a stalagmite: in year n it gains \displaystyle \frac{3n}{4n+5} millimeters of height. Its height after n years is the partial sum S_n of the series \displaystyle \sum_{n=1}^{\infty} \frac{3n}{4n+5}, and no tidy formula for S_n exists, so the telescoping of Infinite series and partial sums has nothing to work on. Still, the yearly gains approach \displaystyle \frac{3}{4} millimeter, and a total that keeps receiving about \displaystyle \frac{3}{4} more each year cannot settle on any height.

That observation, made exact, is the first convergence test. It asks only for the limit of the terms, a question about a sequence (Convergence of sequences), and it works in one direction only: it can rule out a sum, never certify one.

Definition

In plain words: if the terms of a series do not shrink to 0, the series has no sum.

The nth term test for divergence (also called the divergence test). For a series \sum_{n=1}^{\infty} a_n: if \lim_{n \to \infty} a_n does not exist, or exists and is not 0, then the series diverges.

“Does not tend to 0” covers three cases: the terms tend to a nonzero number, the terms tend to \infty or -\infty, or the terms have no limit at all.

If \lim_{n \to \infty} a_n = 0, the test gives no conclusion. The series may converge or diverge, and another argument has to decide.

Intuition

Partial sums move in steps, and the steps are the terms: S_n - S_{n-1} = a_n. A sequence that settles on a number S eventually stays inside any narrow band around S, and a sequence trapped in a band of width 2\varepsilon cannot take a step longer than 2\varepsilon. So the steps must shrink to 0. Turned around: if steps of length at least some fixed c > 0 keep coming, the partial sums never stay inside a band narrower than c, and they cannot settle.

Figure 1 shows Iroda’s yearly gains. From year 3 on each one is at least \displaystyle \frac{1}{2}, so the height climbs by at least \displaystyle \frac{1}{2} millimeter a year and passes every bound.

Figure 1Iroda’s yearly gains \displaystyle a_n = \frac{3n}{4n+5} rise toward \displaystyle \frac{3}{4}. From n = 3 on, every gain is at least \displaystyle \frac{1}{2}, so the total climbs by at least \displaystyle \frac{1}{2} every year.

The steps need not all point the same way. In Figure 2 the terms of \displaystyle \sum \sin\frac{n\pi}{2} run 1, 0, -1, 0 over and over, and the partial sums hop between 1 and 0 forever. They stay bounded, yet a step of length 1 comes every other time, so they never settle either.

Figure 2The partial sums of \displaystyle \sum \sin\frac{n\pi}{2} hop between 1 and 0. Steps of length 1 keep coming, so the partial sums never settle.

Results

The terms of a convergent series tend to 0. If \sum_{n=1}^{\infty} a_n converges, then \lim_{n \to \infty} a_n = 0.

The nth term test. If \lim_{n \to \infty} a_n \ne 0, or the limit does not exist, then \sum_{n=1}^{\infty} a_n diverges.

Terms that stay away from 0. If there is a number c > 0 with \lvert a_n \rvert \ge c for infinitely many n, the series diverges. (This happens exactly when a_n does not tend to 0.)

A positive limit forces divergence to infinity. If a_n \to L with L > 0, or a_n \to \infty, then \sum_{n=1}^{\infty} a_n = \infty. If a_n \to L < 0 or a_n \to -\infty, the partial sums tend to -\infty.

The converse is false. Terms tending to 0 do not make a series converge. The harmonic series and \displaystyle \sum \ln\frac{n+1}{n}, both in Infinite series and partial sums, have terms tending to 0 and diverge to infinity.

Why it is true

The theorem. Suppose S_n \to S. The sequence S_1, S_2, S_3, \ldots shifted by one place, S_{n-1}, has the same limit S, because a limit depends only on the far end of a sequence. For n \ge 2, a_n = S_n - S_{n-1}, so the difference law for limits of sequences gives

\lim_{n \to \infty} a_n = S - S = 0

The test is the contrapositive: if a_n does not tend to 0, the series cannot converge.

Terms that stay away from 0. If the series converged, a_n \to 0, so (taking \varepsilon = c in the definition of the limit) there would be an N with \lvert a_n \rvert < c for every n \ge N. That leaves only finitely many n with \lvert a_n \rvert \ge c, a contradiction.

A positive limit. If a_n \to L > 0, take \displaystyle \varepsilon = \frac{L}{2}: there is an N with \displaystyle a_n > \frac{L}{2} for all n \ge N. Then for n \ge N,

S_n \ge S_{N-1} + (n - N + 1)\frac{L}{2}

which grows without bound, and S_n increases from n = N on, so S_n \to \infty. If a_n \to \infty, then a_n > 1 from some point on and the same argument runs with 1 in place of \displaystyle \frac{L}{2}. For a negative limit, apply this to the terms -a_n, whose partial sums are -S_n.

Worked examples

Example 1

Iroda’s stalagmite

In year n the stalagmite gains \dfrac{3n}{4n+5} millimeters. Show that its height grows without bound. Then use bounds on the terms to say how many years it needs at least before it is taller than 10 millimeters, and a number of years that is certainly enough.

Solution

Divide the top and bottom by n: a_n = \dfrac{3}{4 + 5/n}. Since \displaystyle \frac{5}{n} \to 0, the quotient law gives \displaystyle a_n \to \frac{3}{4}. The limit is not 0, so the series diverges by the nth term test, and since the limit is positive, the height tends to infinity.

At least. Every gain is below \displaystyle \frac{3}{4}, because 4 \cdot 3n = 12n < 12n + 15 = 3(4n + 5). So \displaystyle S_n < \frac{3n}{4}, and S_n > 10 forces \displaystyle \frac{3n}{4} > 10, that is \displaystyle n > \frac{40}{3}. At least 14 years are needed.

Enough. \displaystyle a_n \ge \frac{1}{2} exactly when 6n \ge 4n + 5, that is \displaystyle n \ge \frac{5}{2}, so every gain from year 3 on is at least \displaystyle \frac{1}{2} (Figure 1). With \displaystyle a_1 + a_2 = \frac{1}{3} + \frac{6}{13} = \frac{31}{39},

S_n \ge \frac{31}{39} + \frac{n - 2}{2}

which exceeds 10 once \displaystyle \frac{n-2}{2} > \frac{359}{39}, that is n > 20.41. So 21 years are certainly enough.

Adding the gains directly gives S_{16} \approx 9.84 and S_{17} \approx 10.5387: the stalagmite first passes 10 millimeters in year 17, between the two bounds.

Example 2

Two trigonometric terms

Apply the test to \sum_{n=1}^{\infty} \cos\dfrac{1}{n} and to \sum_{n=1}^{\infty} n\sin\dfrac{1}{n}.

Solution

As n \to \infty, \displaystyle \frac{1}{n} \to 0, and cosine is continuous at 0 (Continuity), so \displaystyle \cos\frac{1}{n} \to \cos 0 = 1. The limit is not 0: the first series diverges, to infinity, since its limit is positive.

For the second, put \displaystyle x = \frac{1}{n}: then \displaystyle n\sin\frac{1}{n} = \frac{\sin x}{x}, and x \to 0^+ as n \to \infty. Since \displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1 (The squeeze theorem and the limit of sine x over x), and a function’s limit as x \to 0^+ is also the limit along the particular values \displaystyle x = \frac{1}{n}, the terms tend to 1. The second series diverges to infinity too.

Without the factor n, the terms \displaystyle \sin\frac{1}{n} tend to 0 and the test says nothing about \displaystyle \sum \sin\frac{1}{n}.

Example 3

Limits known from earlier work

Apply the test to \sum_{n=1}^{\infty} \left(1 + \dfrac{1}{n}\right)^n and to \sum_{n=1}^{\infty} \arctan n.

Solution

The terms \displaystyle \left(1 + \frac{1}{n}\right)^n tend to e \approx 2.718 (The number e and the natural exponential function). The limit is not 0, so the first series diverges. In fact each term is at least 2: by the binomial theorem it is \displaystyle 1 + n \cdot \frac{1}{n} plus terms that are never negative. So S_n \ge 2n.

As x \to \infty, \displaystyle \arctan x \to \frac{\pi}{2}, the horizontal asymptote of the arctangent (Limits at infinity and horizontal asymptotes). If f(x) \to L as x \to \infty, then f(n) \to L along the integers, so \displaystyle \arctan n \to \frac{\pi}{2} \ne 0. The second series diverges to infinity.

Example 4

Terms with no limit

Apply the test to \sum_{n=1}^{\infty} \sin\dfrac{n\pi}{2}.

Solution

The terms repeat with period 4: \displaystyle \sin\frac{\pi}{2} = 1, \sin\pi = 0, \displaystyle \sin\frac{3\pi}{2} = -1, \sin 2\pi = 0, then 1, 0, -1, 0 again. Along n = 1, 5, 9, \ldots every term is 1; along the even n every term is 0. A convergent sequence has every subsequence tending to its limit, and these two subsequences tend to different numbers, so \displaystyle \lim \sin\frac{n\pi}{2} does not exist. By the nth term test the series diverges. (The form with c = 1 applies too: \lvert a_n \rvert = 1 for every odd n.)

The partial sums are 1, 1, 0, 0, 1, 1, 0, 0, \ldots (Figure 2). This series diverges without tending to \infty or -\infty. It is not a geometric series: no fixed ratio carries a_2 = 0 to a_3 = -1.

Example 5

Terms that tend to 0: two series, two verdicts

Apply the test to \sum_{n=1}^{\infty} \dfrac{n}{n^2 + 1} and \sum_{n=1}^{\infty} \dfrac{2}{n(n+2)}, then decide each series another way.

Solution

Dividing by n^2, \displaystyle \frac{n}{n^2+1} = \frac{1/n}{1 + 1/n^2} \to 0, and \displaystyle \frac{2}{n(n+2)} \to 0 because its bottom grows without bound. Both limits are 0, so the test gives no conclusion for either.

The first diverges. For n \ge 1, n^2 + 1 \le 2n^2, so \displaystyle \frac{n}{n^2 + 1} \ge \frac{n}{2n^2} = \frac{1}{2n}. Its partial sums are therefore at least \displaystyle \frac{1}{2}H_n, half the harmonic partial sums, which grow without bound (Infinite series and partial sums).

The second converges. Since \displaystyle \frac{1}{n} - \frac{1}{n+2} = \frac{(n + 2) - n}{n(n+2)} = \frac{2}{n(n+2)}, the series telescopes with a shift of 2:

S_n = 1 + \frac{1}{2} - \frac{1}{n+1} - \frac{1}{n+2} \to \frac{3}{2}

The test treats the two series the same way; their behavior is opposite.

Example 6

Where the test speaks: a parameter

For which real numbers c does the nth term test settle \sum_{n=1}^{\infty} \dfrac{n^c}{n^2 + 1}?

Solution

Write a_n = n^{c-2} \cdot \dfrac{n^2}{n^2 + 1}. The second factor tends to 1.

If c > 2, then n^{c-2} \to \infty, so a_n \to \infty and the series diverges.
If c = 2, then a_n \to 1 and the series diverges.
If c < 2, then \displaystyle n^{c-2} = \frac{1}{n^{2-c}} \to 0, so a_n \to 0 and the test says nothing.

Both outcomes occur when c < 2. At c = 1 the series is the first one in Example 5, which diverges. At c = 0 the terms are positive and, for n \ge 2, \displaystyle \frac{1}{n^2 + 1} < \frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}. That last series telescopes, so

S_n < \frac{1}{2} + \left(1 - \frac{1}{n}\right) < \frac{3}{2}

The partial sums increase and stay below \displaystyle \frac{3}{2}, so the series converges.

Comparison with these two cases settles more. For 1 \le c < 2, n^c \ge n, so each term is at least the matching term at c = 1, and the series diverges by Example 5. For c \le 0, n^c \le 1, so each term is at most the matching term at c = 0; the partial sums increase and stay below \displaystyle \frac{3}{2}, and the series converges. The remaining range 0 < c < 1 also converges, so the dividing line is c = 1. That last step comes from The integral test and p series and The direct and limit comparison tests and is stated here without proof.

Example 7

Kamola’s convergent series

Kamola knows only that \sum_{n=1}^{\infty} a_n converges and that no a_n equals -1 or 3. Find \lim_{n \to \infty} \dfrac{2a_n + 5}{3 - a_n}, and decide whether \sum_{n=1}^{\infty} \dfrac{1}{1 + a_n} converges.

Solution

The series converges, so its terms tend to 0. By the limit laws the top tends to 5 and the bottom to 3 \ne 0, so

\lim_{n \to \infty} \frac{2a_n + 5}{3 - a_n} = \frac{5}{3}

Likewise \displaystyle \frac{1}{1 + a_n} \to 1 \ne 0, so \displaystyle \sum \frac{1}{1 + a_n} diverges by the nth term test. Here the theorem is used forward, as a fact a convergent series hands over, and then the test is used on a new series.

Boundaries

The test only proves divergence. \lim a_n = 0 is necessary for convergence and never sufficient (Examples 5 and 6). A limit of 0 ends the test, not the question.
It says nothing about how a series diverges. When the test applies, the partial sums may tend to \infty (Example 1) or oscillate (Example 4). Only the sign of a nonzero limit adds that information.
It gives no sum. Even for a series known to converge, the limit of the terms is 0 and carries no information about the value.
The terms, not a function through them. If f(x) \to L as x \to \infty, then f(n) \to L. The reverse fails: \sin(n\pi) = 0 for every integer n, so \sum \sin(n\pi) converges to 0, although \lim_{x \to \infty} \sin(\pi x) does not exist. A function with no limit at infinity does not show that the terms have none.
Finitely many terms do not matter. The limit of a_n depends only on the far end of the sequence, so the test reads \sum_{n=m}^{\infty} a_n for any starting index m the same way.

Where this goes next

The test is the first look at any series. When the terms do tend to 0, the Convergence tests take over: comparison with a known series (The direct and limit comparison tests), the integral test and p-series (The integral test and p series), and the The ratio and root tests. For a geometric series the case \lvert r \rvert \ge 1 in Infinite geometric series is this test at work. In power series, a series that converges on an interval often fails exactly at an endpoint where its terms stop tending to 0, which is one of the checks in Power series and the interval of convergence.

Exercises

Applying the test

Find the limit of the terms, if it exists, and say what the nth term test concludes.

\sum_{n=1}^{\infty} \dfrac{5n^2 - 1}{2n^2 + 7n}
\displaystyle \sum_{n=1}^{\infty} \dfrac{n}{\sqrt{9n^2 + 4}}
\sum_{n=1}^{\infty} \dfrac{2^n}{2^n + 3}
\displaystyle \sum_{n=1}^{\infty} \left(\sqrt{n^2 + n} - n\right)
\sum_{n=1}^{\infty} \left(\dfrac{n}{n+2}\right)^n
\sum_{n=2}^{\infty} \dfrac{\ln n}{\ln(3n)}
\sum_{n=1}^{\infty} \cos\dfrac{2n\pi}{3}
\sum_{n=1}^{\infty} (-1)^n \dfrac{n^2}{n^2 + 1}
\sum_{n=1}^{\infty} \dfrac{e^n}{n^3}
\sum_{n=1}^{\infty} n\left(e^{1/n} - 1\right)

When the terms tend to 0

What does the test say about \sum_{n=1}^{\infty} n\left(1 - \cos\dfrac{1}{n}\right)?
Show that the test says nothing about \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{\sqrt{n}}, then show the series diverges by proving \displaystyle S_n \ge \sqrt{n}.
For which real numbers b does the test show that \sum_{n=1}^{\infty} \dfrac{bn^2 + 1}{3n^2 + n} diverges?

Convergent series, and a setting

Suppose \sum_{n=1}^{\infty} a_n converges and every a_n > 0. Find \lim_{n \to \infty} \dfrac{a_n + 4}{2a_n + 1} and \displaystyle \lim_{n \to \infty} \sqrt{9 + a_n}.
Suppose \sum_{n=1}^{\infty} a_n converges. Decide whether \sum_{n=1}^{\infty} e^{a_n} and \sum_{n=1}^{\infty} (2 - a_n) converge.
Suppose \sum_{n=1}^{\infty} a_n converges and every a_n > 0. Show that \sum_{n=1}^{\infty} \dfrac{1}{a_n} diverges.
Yulduz models a second stalagmite that gains \dfrac{5n}{8n + 3} millimeters in year n. Show that its height grows without bound, use \displaystyle a_n < \frac{5}{8} to find the least number of years it needs before it is taller than 6 millimeters, and find by adding the gains the first year it is taller.

Reasoning

Prove that if a_n \to L with L < 0, then the partial sums of \sum a_n tend to -\infty.
Kamola claims that every series whose terms tend to 0 converges. Give a counterexample other than the harmonic series, with a reason.
Explain why the nth term test can never show that a series converges.
Show that \sum_{n=1}^{\infty} \sin n diverges. (Suppose \sin n \to 0 and use the addition formula for \sin(n + 1).)
Show that if \sum a_n converges, then a_n + a_{n+1} \to 0. Then give a divergent series for which a_n + a_{n+1} \to 0.

Answers

Limit \displaystyle \frac{5}{2}; diverges. Dividing by n^2, the terms are \displaystyle \frac{5 - 1/n^2}{2 + 7/n}.
Limit \displaystyle \frac{1}{3}; diverges. \displaystyle \frac{n}{\sqrt{9n^2 + 4}} = \frac{1}{\sqrt{9 + 4/n^2}}.
Limit 1; diverges. \displaystyle \frac{2^n}{2^n + 3} = \frac{1}{1 + 3/2^n} and \displaystyle \frac{3}{2^n} \to 0.
Limit \displaystyle \frac{1}{2}; diverges. Multiplying by the conjugate, the term is \displaystyle \frac{n}{\sqrt{n^2 + n} + n}. Dividing top and bottom by n gives \displaystyle \frac{1}{\sqrt{1 + 1/n} + 1}, which tends to \displaystyle \frac{1}{1 + 1}.
Limit e^{-2} \approx 0.1353; diverges. The term is \displaystyle \frac{1}{(1 + 2/n)^n}, and with \displaystyle x = \frac{2}{n}, \displaystyle n\ln\left(1 + \frac{2}{n}\right) = 2 \cdot \frac{\ln(1+x)}{x} \to 2, so \displaystyle \left(1 + \frac{2}{n}\right)^n \to e^2.
Limit 1; diverges. \displaystyle \frac{\ln n}{\ln 3 + \ln n} = \frac{1}{1 + \ln 3/\ln n} and \ln n \to \infty.
No limit; diverges. The terms run \displaystyle -\frac{1}{2}, -\frac{1}{2}, 1 and repeat, so n = 3k gives 1 and n = 3k + 1 gives \displaystyle -\frac{1}{2}.
No limit; diverges. Even n give \displaystyle \frac{n^2}{n^2 + 1} \to 1, odd n give \displaystyle -\frac{n^2}{n^2 + 1} \to -1.
The terms tend to \infty; diverges to infinity. An exponential outgrows every power (Comparing growth rates at infinity).
Limit 1; diverges. With \displaystyle x = \frac{1}{n}, the term is \displaystyle \frac{e^x - 1}{x}, which tends to the derivative of e^x at 0, namely 1.
Nothing. With \displaystyle x = \frac{1}{n}, \displaystyle \frac{1 - \cos x}{x} = \frac{\sin x}{x} \cdot \frac{\sin x}{1 + \cos x} \to 1 \cdot \frac{0}{2} = 0. The series still diverges: in the same way \displaystyle \frac{1 - \cos x}{x^2} equals \displaystyle \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x}, which tends to \displaystyle \frac{1}{2}, so from some N on the term, which equals \displaystyle \frac{1}{n} \cdot \frac{1 - \cos x}{x^2}, is at least \displaystyle \frac{1}{4n}, and the partial sums grow at least as fast as \displaystyle \frac{1}{4} of the harmonic partial sums.
Diverges. \displaystyle \frac{1}{\sqrt{n}} \to 0, so the test is silent. Each of the first n terms is at least \displaystyle \frac{1}{\sqrt{n}}, so \displaystyle S_n \ge n \cdot \frac{1}{\sqrt{n}} = \sqrt{n} \to \infty.
Every b \ne 0. The terms tend to \displaystyle \frac{b}{3}. For b = 0 the terms \displaystyle \frac{1}{3n^2 + n} tend to 0 and the test says nothing.
4 and 3. a_n \to 0, so the quotient tends to \displaystyle \frac{0 + 4}{0 + 1}, and \displaystyle \sqrt{9 + a_n} \to \sqrt{9} by continuity of the square root.
Both diverge. a_n \to 0, so e^{a_n} \to e^0 = 1 and 2 - a_n \to 2, neither 0.
a_n \to 0 with a_n > 0. For any M > 0, eventually \displaystyle a_n < \frac{1}{M}, so \displaystyle \frac{1}{a_n} > M. Thus \displaystyle \frac{1}{a_n} \to \infty, and the test gives divergence.
\displaystyle a_n \to \frac{5}{8} \ne 0, positive, so the height tends to infinity. \displaystyle S_n < \frac{5n}{8}, so being taller than 6 millimeters needs \displaystyle n > \frac{48}{5} = 9.6: at least 10 years. Adding, S_{10} \approx 5.6701 and S_{11} \approx 6.2745, so year 11.
The terms -a_n tend to -L > 0, so by the positive-limit result their partial sums -S_n tend to \infty, which means S_n \to -\infty.
\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} (Exercise 12): its terms tend to 0, and its partial sums satisfy \displaystyle S_n \ge \sqrt{n}, so they grow without bound. (\displaystyle \sum \frac{n}{n^2+1} from Example 5 and \displaystyle \sum \ln\frac{n+1}{n} also work.)
The theorem runs one way: convergence forces a_n \to 0. A limit of 0 is shared by convergent series and divergent ones (Example 5), so it cannot separate them; the test only detects a limit other than 0, which rules convergence out.
Suppose \sin n \to 0. Then \sin(n + 1) \to 0 too, and \sin(n+1) = \sin n\cos 1 + \cos n\sin 1, so \cos n \sin 1 \to 0. Since \sin 1 \ne 0, \cos n \to 0. Then \sin^2 n + \cos^2 n \to 0, but it equals 1 for every n. So \sin n does not tend to 0, and the series diverges.
a_n \to 0, and the shifted sequence a_{n+1} \to 0, so the sum law gives a_n + a_{n+1} \to 0. For a_n = (-1)^n, every a_n + a_{n+1} = 0, but \lvert a_n \rvert = 1 for all n, so \sum (-1)^n diverges.