Mathematics guide, Unit 10, step 29 of 33Previous step Next step

Step 29: Series and Convergence Tests

By the EconTutorials team. 11-minute read with 6 practice questions.

SyllabusAP BC 10.1–10.10

In 30 seconds

AP Calculus BC only. This step is AP Unit 10 (topics 10.1 to 10.10). It is not in AP Calculus AB, Cambridge 9709 or IB AA, although the geometric series of Step 6 is common to all of them.
An infinite series $ \sum a_n $ converges if its partial sums $ S_n = a_1 + a_2 + \cdots + a_n $ approach a finite limit. If the terms do not tend to 0, it diverges; if they do, you still need a test.
The toolkit: geometric series, the $ n $th-term test, the integral test, $ p $-series, direct and limit comparison, the alternating series test and the ratio test. A series is absolutely convergent if $ \sum |a_n| $ converges, and conditionally convergent if $ \sum a_n $ converges but $ \sum |a_n| $ does not.
Exam point: name the test and check its conditions in words. For an alternating series, the error after $ n $ terms is at most the first term left out.

Video lesson coming soonA short animated lesson for this step will appear here.

Sequences, series and partial sums

AP Calculus BC only. A sequence is a list of numbers $ a_1, a_2, a_3, \ldots $. It converges if its terms settle on a limit: for example $ a_n = \dfrac{3n + 1}{2n-5} $ tends to $ \tfrac{3}{2} $, because for large $ n $ the top is about $ 3n $ and the bottom about $ 2n $. A series is what you get when you try to add the terms of a sequence, $ a_1 + a_2 + a_3 + \cdots $.

You can never finish adding infinitely many numbers, but you can watch the running totals.

Key termPartial sum: $ S_n = a_1 + a_2 + \cdots + a_n $. The series $ \sum_{n=1}^{\infty} a_n $ converges to $ S $ if $ S_n \to S $, a finite number. Otherwise it diverges.

Keep the two ideas apart. The sequence $ a_n = \tfrac{1}{n} $ converges (to 0), but the series $ 1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots $ diverges, as you will see below.

The most important series of all is the geometric series from Step 6. With first term $ a $ and common ratio $ r $:

\[ \sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \cdots = \frac{a}{1-r} \quad \text{if } |r| < 1, \]

and it diverges if $ |r| \ge 1 $. The reason is that $ S_n = \dfrac{a(1-r^n)}{1-r} $, and $ r^n \to 0 $ exactly when $ |r| < 1 $.

A few other series can be summed exactly because their partial sums collapse. These are called telescoping series, and partial fractions are the usual way in.

Worked example

Find the sum of (a) $ \displaystyle\sum_{n=1}^{\infty} \frac{2}{n(n + 2)} $ and (b) $ \displaystyle\sum_{n=1}^{\infty} 4\left(-\frac{1}{3}\right)^{n-1} $.

Step 1(a) Partial fractions: $ \dfrac{2}{n(n + 2)} = \dfrac{1}{n}-\dfrac{1}{n + 2} $.

Step 2Write out $ S_n $: $ \big(1-\tfrac{1}{3}\big) + \big(\tfrac{1}{2}-\tfrac{1}{4}\big) + \big(\tfrac{1}{3}-\tfrac{1}{5}\big) + \cdots + \big(\tfrac{1}{n}-\tfrac{1}{n + 2}\big) $. Every negative part cancels with a positive part two brackets later.

Step 3What survives is $ S_n = 1 + \tfrac{1}{2}-\tfrac{1}{n + 1}-\tfrac{1}{n + 2} $, which tends to $ \tfrac{3}{2} $. So the sum is $ \tfrac{3}{2} $.

Step 4(b) This is geometric with $ a = 4 $ and $ r = -\tfrac{1}{3} $. Since $ |r| < 1 $, the sum is $ \dfrac{4}{1 + \tfrac{1}{3}} = 3 $.

In (b) the first term is the $ n = 1 $ term, $ 4 $. Always check which term the sum starts from before using $ \dfrac{a}{1-r} $.

Here is an economics use of the geometric series. A share pays a dividend of $3 next year, and the dividend grows by 4% a year after that. If investors discount future money at 10% a year, the share’s present value is

\[ \frac{3}{1.1} + \frac{3(1.04)}{1.1^2} + \frac{3(1.04)^2}{1.1^3} + \cdots \]

This is geometric, with first term $ \tfrac{3}{1.1} $ and ratio $ \tfrac{1.04}{1.1} < 1 $, so its sum is $ \dfrac{3/1.1}{1-1.04/1.1} = \dfrac{3}{0.10-0.04} = \$50 $.

This is the Gordon growth formula, $ \dfrac{D}{i-g} $. It only works when the growth rate $ g $ is below the discount rate $ i $. If $ g \ge i $, the ratio is at least 1, the series diverges, and the model gives no finite price.

The nth-term test, the integral test and p-series

AP Calculus BC only. Most series cannot be summed exactly. A convergence test answers an easier question: does a sum exist at all? The first test is a quick check for divergence.

Key termnth-term test for divergence: if $ \displaystyle\lim_{n \to \infty} a_n \neq 0 $, or the limit does not exist, then $ \sum a_n $ diverges.

It works because if $ S_n \to S $ then $ a_n = S_n-S_{n-1} \to S-S = 0 $. For example $ \sum \dfrac{2n^2}{n^2 + 1} $ diverges, because its terms approach 2: you are adding nearly 2 for ever.

Common mistakeUsing the test backwards. If $ a_n \to 0 $, the $ n $th-term test tells you nothing. The harmonic series $ 1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \cdots $ has terms tending to 0, yet it diverges. Group it as $ 1 + \tfrac{1}{2} + \big(\tfrac{1}{3} + \tfrac{1}{4}\big) + \big(\tfrac{1}{5} + \cdots + \tfrac{1}{8}\big) + \cdots $: each bracket is more than $ \tfrac{1}{2} $, so the total passes every number eventually.

Partial sums Sₙ of Σ 1/n and Σ 1/n²

Σ 1/n (harmonic): keeps climbingΣ 1/n²: levels off at π²/6 ≈ 1.645

Figure 1: both series have terms that tend to 0. The harmonic partial sums grow like ln n and pass every number eventually; the partial sums of Σ 1/n² settle on π²/6.

When the terms come from a function you can integrate, compare the sum with an integral.

Key termIntegral test: if $ f $ is continuous, positive and decreasing for $ x \ge k $, and $ a_n = f(n) $, then $ \sum_{n=k}^{\infty} a_n $ and $ \displaystyle\int_k^{\infty} f(x)\,dx $ either both converge or both diverge.

Figure 2 shows why. Rectangles of width 1 and heights $ a_2, a_3, a_4, \ldots $ fit under the curve, so if the area under the curve is finite, the partial sums are capped. Shift the rectangles one place to the left and they sit above the curve, so if the area is infinite, so is the sum. The integral and the sum are usually different numbers; they only share a fate.

Integral test: rectangles of height aₙ = 1/n² under y = 1/x²

y = 1/x²rectangles a₂ + a₃ + … (width 1)

Figure 2: the rectangle on [n − 1, n] has height f(n) and lies under the curve, so the partial sum a₂ + a₃ + … up to any n is less than the area under y = 1/x² from 1 to n, which is never more than 1. The partial sums are capped, so Σ 1/n² converges.

Applying the test to $ f(x) = x^{-p} $ and using the improper integrals of Step 23 gives the benchmark family you will compare almost everything with.

\[ \sum_{n=1}^{\infty} \frac{1}{n^p} \ \text{ converges if } p > 1 \text{ and diverges if } p \le 1. \]

So $ \sum \tfrac{1}{n^2} $ and $ \sum \tfrac{1}{n\sqrt{n}} $ converge, while $ \sum \tfrac{1}{n} $ and $ \sum \tfrac{1}{\sqrt{n}} $ diverge.

Worked example

Use the integral test to decide whether $ \displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2} $ converges.

Step 1Let $ f(x) = \dfrac{1}{x(\ln x)^2} $. For $ x \ge 2 $ it is continuous and positive, and it is decreasing because both $ x $ and $ (\ln x)^2 $ increase.

Step 2Substitute $ u = \ln x $, $ du = \dfrac{dx}{x} $: $ \displaystyle\int \frac{dx}{x(\ln x)^2} = \int u^{-2}\,du = -\frac{1}{\ln x} + C $.

Step 3$ \displaystyle\int_2^{\infty} f(x)\,dx = \lim_{b \to \infty}\Big[-\frac{1}{\ln x}\Big]_2^b = \lim_{b \to \infty}\Big(\frac{1}{\ln 2}-\frac{1}{\ln b}\Big) = \frac{1}{\ln 2} $.

Step 4The integral is finite, so the series converges by the integral test.

The series converges, but its sum is not $ \tfrac{1}{\ln 2} \approx 1.44 $. The test says nothing about the value of the sum.

Comparison tests

AP Calculus BC only. A series that looks like a $ p $-series or a geometric series usually behaves like one. The comparison tests, for series with positive terms, turn “looks like” into a proof.

Test	Condition	Conclusion
Direct comparison	$ 0 \le a_n \le b_n $ for all $ n $	$ \sum b_n $ converges $ \Rightarrow $ $ \sum a_n $ converges; $ \sum a_n $ diverges $ \Rightarrow $ $ \sum b_n $ diverges
Limit comparison	$ a_n, b_n > 0 $ and $ \displaystyle\lim_{n \to \infty} \frac{a_n}{b_n} = c $, with $ 0 < c < \infty $	both converge or both diverge

A pile smaller than a finite pile is finite. And if $ a_n $ is eventually about $ c $ times $ b_n $, the two series must share a fate. To choose $ b_n $, keep only the highest power of $ n $ on the top and on the bottom.

Worked example

Decide whether each series converges: (a) $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{3^n + 2} $ (b) $ \displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2-n + 1} $.

Step 1(a) For every $ n $, $ 3^n + 2 > 3^n $, so $ 0 < \dfrac{1}{3^n + 2} < \left(\dfrac{1}{3}\right)^n $.

Step 2$ \sum \big(\tfrac{1}{3}\big)^n $ is geometric with $ r = \tfrac{1}{3} $, so it converges. By direct comparison, so does (a).

Step 3(b) For large $ n $ the terms behave like $ \dfrac{n^{1/2}}{n^2} = \dfrac{1}{n^{3/2}} $. Take $ b_n = \dfrac{1}{n^{3/2}} $.

Step 4$ \dfrac{a_n}{b_n} = \dfrac{n^{1/2} \cdot n^{3/2}}{n^2-n + 1} = \dfrac{n^2}{n^2-n + 1} \to 1 $, which is finite and positive.

Step 5$ \sum \dfrac{1}{n^{3/2}} $ is a $ p $-series with $ p = \tfrac{3}{2} > 1 $, so it converges, and by limit comparison so does (b).

In (b) a direct comparison would be awkward, because the denominator is smaller than $ n^2 $ and the inequality points the useless way. Limit comparison avoids the problem.

Alternating series and absolute convergence

AP Calculus BC only. When the signs alternate, the terms partly cancel each other, and a series can converge even though its terms shrink slowly.

Key termAlternating series test: if $ b_n > 0 $, the $ b_n $ are decreasing (at least from some point on) and $ b_n \to 0 $, then $ \sum (-1)^{n+1} b_n = b_1-b_2 + b_3-\cdots $ converges.

Each step goes the other way and is shorter than the last, so the partial sums zigzag and close in on the sum from both sides. That gives a bonus: the sum always lies between two neighbouring partial sums. So the alternating series error bound is

\[ |S-S_n| \le b_{n+1}, \]

the size of the first term left out, and the error has the same sign as that term.

Partial sums of 1 − 1/3 + 1/5 − 1/7 + …

odd n: above the sumeven n: below the sum

Figure 3: the partial sums zigzag around the limit π/4, overshooting by less each time. Each gap |S − Sₙ| is smaller than the next term, 1/(2n + 1).

Key termAbsolute and conditional convergence: if $ \sum |a_n| $ converges, then $ \sum a_n $ converges too, and it is absolutely convergent. If $ \sum a_n $ converges but $ \sum |a_n| $ diverges, it is conditionally convergent.

Worked example

(a) Show that $ \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}} $ converges conditionally. (b) For $ \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3} $, estimate the sum with $ S_3 $, bound the error, and find how many terms guarantee an error below 0.001.

Step 1(a) The terms $ b_n = \dfrac{1}{\sqrt{n}} $ are positive, decreasing and tend to 0, so the series converges by the alternating series test.

Step 2But $ \sum \dfrac{1}{\sqrt{n}} $ is a $ p $-series with $ p = \tfrac{1}{2} \le 1 $, so it diverges. The convergence is conditional.

Step 3(b) $ S_3 = 1-\tfrac{1}{8} + \tfrac{1}{27} = \tfrac{197}{216} \approx 0.9120 $. The error is at most $ b_4 = \tfrac{1}{64} \approx 0.0156 $.

Step 4The first omitted term, $ -\tfrac{1}{64} $, is negative, so $ S_3 $ is an overestimate.

Step 5For an error below 0.001 we need $ \dfrac{1}{(n + 1)^3} < 0.001 $, so $ (n + 1)^3 > 1000 $ and $ n + 1 > 10 $. Ten terms are enough: the bound is then $ \tfrac{1}{1331} $.

The true sum is about 0.9015, which is indeed below $ S_3 $ by less than $ \tfrac{1}{64} $. Series (b) also converges absolutely, since $ \sum \tfrac{1}{n^3} $ is a $ p $-series with $ p = 3 $.

The ratio test, and choosing a test

AP Calculus BC only. When the terms contain factorials or powers such as $ 3^n $, compare each term with the one before it.

Key termRatio test: let $ L = \displaystyle\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| $. If $ L < 1 $, $ \sum a_n $ converges absolutely. If $ L > 1 $ (including $ L = \infty $), it diverges. If $ L = 1 $, the test gives no conclusion.

If neighbouring terms have a ratio close to $ L $, the tail of the series behaves like a geometric series with ratio $ L $. At $ L = 1 $ anything can happen: $ \sum \tfrac{1}{n} $ diverges and $ \sum \tfrac{1}{n^2} $ converges, yet both give $ L = 1 $.

Worked example

Use the ratio test on (a) $ \displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n} $ and (b) $ \displaystyle\sum_{n=1}^{\infty} \frac{n!}{5^n} $.

Step 1(a) $ \dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)^2}{2^{n+1}} \cdot \dfrac{2^n}{n^2} = \dfrac{1}{2}\left(\dfrac{n + 1}{n}\right)^2 $.

Step 2As $ n \to \infty $, $ \dfrac{n + 1}{n} \to 1 $, so $ L = \tfrac{1}{2} < 1 $. The series converges absolutely.

Step 3(b) $ \dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)!}{5^{n+1}} \cdot \dfrac{5^n}{n!} = \dfrac{n + 1}{5} $.

Step 4This tends to infinity, so $ L > 1 $ and the series diverges.

In (b) the factorial eventually beats the power of 5, so the terms grow without limit; the $ n $th-term test would also show divergence.

With practice, the form of the terms tells you which test to try first.

If the series looks like	Try
terms that do not tend to 0	$ n $th-term test: diverges
$ ar^n $	geometric: converges if and only if $ \|r\| < 1 $
$ \dfrac{1}{n^p} $, or a ratio of powers of $ n $	$ p $-series, then direct or limit comparison
signs that alternate	alternating series test, then test $ \sum \|a_n\| $
factorials, or $ n $ in an exponent	ratio test
$ f(n) $ with an easy integral	integral test

Exam tipA bare “converges” earns little credit on an AP free-response question. Name the test and check every condition in words, for example: “the terms $ \tfrac{1}{\sqrt{n}} $ are positive, decreasing and have limit 0, so the series converges by the alternating series test.” For limit comparison, state the limit and that it is finite and positive.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find the sum of $ \displaystyle\sum_{n=0}^{\infty} 6\left(\frac{1}{4}\right)^n $. Show worked answer

Geometric, with first term $ a = 6 $ (the $ n = 0 $ term) and ratio $ r = \tfrac{1}{4} $.
$ |r| < 1 $, so the series converges.
Sum $ = \dfrac{6}{1-\tfrac{1}{4}} = \dfrac{6}{3/4} = 8 $.

EasyQ2. Does $ \displaystyle\sum_{n=1}^{\infty} \frac{n + 1}{3n} $ converge? Show worked answer

$ \displaystyle\lim_{n \to \infty} \frac{n + 1}{3n} = \frac{1}{3} \neq 0 $.
So the series diverges by the $ n $th-term test.

MediumQ3. Decide whether $ \displaystyle\sum_{n=1}^{\infty} \frac{2n + 1}{n^3 + 5} $ converges. Show worked answer

For large $ n $ the terms behave like $ \dfrac{2n}{n^3} = \dfrac{2}{n^2} $. Compare with $ b_n = \dfrac{1}{n^2} $.
$ \dfrac{a_n}{b_n} = \dfrac{(2n + 1)n^2}{n^3 + 5} = \dfrac{2n^3 + n^2}{n^3 + 5} \to 2 $, finite and positive.
$ \sum \dfrac{1}{n^2} $ converges ($ p = 2 > 1 $), so the series converges by limit comparison.

MediumQ4. Use the ratio test to decide whether $ \displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n \cdot 2^n} $ converges. Show worked answer

$ \dfrac{a_{n+1}}{a_n} = \dfrac{3^{n+1}}{(n + 1)2^{n+1}} \cdot \dfrac{n \cdot 2^n}{3^n} = \dfrac{3}{2} \cdot \dfrac{n}{n + 1} $.
As $ n \to \infty $ this tends to $ L = \tfrac{3}{2} > 1 $.
So the series diverges by the ratio test.

MediumQ5. Is $ \displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n n}{n^2 + 1} $ absolutely convergent, conditionally convergent or divergent? Show worked answer

Let $ b_n = \dfrac{n}{n^2 + 1} > 0 $. With $ f(x) = \dfrac{x}{x^2 + 1} $, $ f'(x) = \dfrac{1-x^2}{(x^2 + 1)^2} \le 0 $ for $ x \ge 1 $, so $ b_n $ is decreasing.
$ b_n \to 0 $, so the series converges by the alternating series test.
Absolute values: $ \sum \dfrac{n}{n^2 + 1} $. Compare with $ \dfrac{1}{n} $: the ratio $ \dfrac{n^2}{n^2 + 1} \to 1 $, so by limit comparison it diverges like the harmonic series.
So the series is conditionally convergent.

ExamQ6. Consider the series $ \displaystyle S = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \cdot 3^n} $. (a) Use the ratio test to show that the series converges absolutely. [2] (b) Find the partial sum $ S_3 $ as a fraction, and use the alternating series error bound to show that $ |S-S_3| < 0.004 $. [3] (c) Is $ S_3 $ greater than or less than $ S $? Give a reason. [1] Show worked answer

(a) $ \left|\dfrac{a_{n+1}}{a_n}\right| = \dfrac{n \cdot 3^n}{(n + 1)3^{n+1}} = \dfrac{1}{3} \cdot \dfrac{n}{n + 1} $. [M1]
This tends to $ \tfrac{1}{3} < 1 $, so the series converges absolutely. [A1]
(b) $ S_3 = \dfrac{1}{3}-\dfrac{1}{18} + \dfrac{1}{81} = \dfrac{54-9 + 2}{162} = \dfrac{47}{162} \approx 0.2901 $. [B1]
The terms $ \dfrac{1}{n \cdot 3^n} $ are positive, decreasing and tend to 0, so the alternating series error bound applies. [M1]
$ |S-S_3| \le \dfrac{1}{4 \cdot 3^4} = \dfrac{1}{324} \approx 0.0031 < 0.004 $. [A1]
(c) The first omitted term, $ -\dfrac{1}{324} $, is negative, so $ S_3 $ is greater than $ S $. [B1]
(In Step 30 you will see that $ S = \ln\tfrac{4}{3} \approx 0.2877 $.)
Total: 6 marks.

Check yourself

1. The series $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{0.9}} $:

A $ p $-series converges only when $ p > 1 $. Here $ p = 0.9 $, so it diverges, even though its terms tend to 0.

2. You know that $ \displaystyle\lim_{n \to \infty} a_n = 0 $. What can you conclude about $ \sum a_n $?

Terms tending to 0 is necessary for convergence but not enough: the harmonic series has $ a_n \to 0 $ and diverges, while $ \sum \tfrac{1}{n^2} $ converges.

3. For $ \displaystyle\sum_{n=1}^{\infty} \frac{5^n}{n!} $, the ratio test gives:

$ \dfrac{a_{n+1}}{a_n} = \dfrac{5}{n + 1} \to 0 $, and $ 0 < 1 $, so the series converges absolutely.

Your score: 0 out of 3

The short version

This step is AP Calculus BC only. A series converges when its partial sums $ S_n $ tend to a finite limit.
Geometric: $ \sum ar^n = \dfrac{a}{1-r} $ when $ |r| < 1 $. $ p $-series: $ \sum \tfrac{1}{n^p} $ converges exactly when $ p > 1 $; the harmonic series diverges.
If $ a_n \not\to 0 $, the series diverges. If $ a_n \to 0 $, use another test: integral, direct or limit comparison, alternating series or ratio.
Absolute convergence ($ \sum |a_n| $ converges) implies convergence. Conditional convergence means $ \sum a_n $ converges but $ \sum |a_n| $ does not.
For an alternating series that passes the test, $ |S-S_n| \le b_{n+1} $, the first term left out.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

If the series looks like	Try
terms that do not tend to 0	\( n \)th-term test: diverges
\( ar^n \)	geometric: converges if and only if \( \|r\| < 1 \)
\( \dfrac{1}{n^p} \), or a ratio of powers of \( n \)	\( p \)-series, then direct or limit comparison
signs that alternate	alternating series test, then test \( \sum \|a_n\| \)
factorials, or \( n \) in an exponent	ratio test
\( f(n) \) with an easy integral	integral test

Step 29: Series and Convergence Tests

Sequences, series and partial sums

The nth-term test, the integral test and p-series

Comparison tests

Alternating series and absolute convergence

The ratio test, and choosing a test

Practise: 6 questions

Check yourself

The short version

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