What it means for a series to converge
Before you start: a series is built from a sequence, so make sure you're comfortable with convergence of sequences.
A series is an infinite sum \(\displaystyle\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots\). You can't add infinitely many numbers directly, so instead you look at the partial sums
\[S_n = a_1 + a_2 + \cdots + a_n\]
The series \(\sum a_n\) converges to \(S\) if its sequence of partial sums \(S_n\) converges to \(S\). Otherwise it diverges.
Zeno again: \(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots\) has partial sums \(\tfrac{1}{2}, \tfrac{3}{4}, \tfrac{7}{8}, \dots\), which converge to 1. So this infinite sum has the finite value 1.
Two benchmark series
Most tests work by comparing your series with one of these two. Learn them by heart.
| Series | Converges when | Diverges when |
|---|---|---|
| Geometric \(\displaystyle\sum_{n=0}^{\infty} a r^n\) | \(|r| < 1\), with sum \(\dfrac{a}{1 - r}\) | \(|r| \ge 1\) |
| \(p\)-series \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^p}\) | \(p > 1\) | \(p \le 1\) |
The case \(p = 1\) is the harmonic series \(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \cdots\), which diverges even though its terms tend to 0. Group the terms: \(\tfrac{1}{3} + \tfrac{1}{4} > \tfrac{1}{2}\), \(\tfrac{1}{5} + \cdots + \tfrac{1}{8} > \tfrac{1}{2}\), and so on. You can collect another \(\tfrac{1}{2}\) as many times as you like, so the sum grows without bound.
The tests, and when to use each
| Test | What it says | Use it when |
|---|---|---|
| \(n\)th-term test | If \(a_n \not\to 0\), the series diverges. | Always check this first. It can only prove divergence. |
| Comparison | If \(0 \le a_n \le b_n\) and \(\sum b_n\) converges, so does \(\sum a_n\). If \(a_n \ge b_n \ge 0\) and \(\sum b_n\) diverges, so does \(\sum a_n\). | \(a_n\) is clearly smaller or larger than a benchmark series. |
| Limit comparison | If \(\dfrac{a_n}{b_n} \to\) a finite non-zero number, then \(\sum a_n\) and \(\sum b_n\) behave the same way. | \(a_n\) is a ratio of powers of \(n\). Take \(b_n = \dfrac{1}{n^{p}}\), where \(p\) = (power in the denominator) − (power in the numerator). |
| D'Alembert's ratio test | Let \(\dfrac{a_{n+1}}{a_n} \to l\). Converges if \(l < 1\), diverges if \(l > 1\), no conclusion if \(l = 1\). | \(a_n\) has factorials or powers like \(2^n\). |
| Cauchy's root test | Let \((a_n)^{1/n} \to l\). Same conclusions as the ratio test. | The whole term is raised to a power of \(n\), like \(\left(\dfrac{n}{n+1}\right)^{n^2}\). |
| Raabe's test | Let \(n\left(\dfrac{a_n}{a_{n+1}} - 1\right) \to k\). Converges if \(k > 1\), diverges if \(k < 1\). | The ratio test gives \(l = 1\). |
| Leibniz's test | For an alternating series \(\sum (-1)^{n+1} u_n\), with \(u_n > 0\): if \(u_n\) decreases and \(u_n \to 0\), the series converges. | The signs alternate, + − + − … |
Absolute and conditional convergence. If \(\sum |a_n|\) converges, the series is absolutely convergent, and it converges. If \(\sum a_n\) converges but \(\sum |a_n|\) does not, it is conditionally convergent.
The first five tests apply to series of positive terms.
Solved examples
Example 1: The nth-term test
EasyTest \(\displaystyle\sum \frac{n}{n + 1}\) for convergence.
- \(a_n = \dfrac{n}{n + 1} \to 1 \ne 0\).
- Terms that don't shrink to zero can't add up to a finite total.
AnswerDiverges, by the \(n\)th-term test.
Example 2: Direct comparison
EasyTest \(\displaystyle\sum \frac{1}{n^2 + 1}\).
- \(0 < \dfrac{1}{n^2 + 1} < \dfrac{1}{n^2}\) for every \(n\).
- \(\sum \dfrac{1}{n^2}\) is a \(p\)-series with \(p = 2 > 1\), so it converges.
AnswerConverges, by comparison.
Example 3: Limit comparison
MediumTest \(\displaystyle\sum \frac{2n + 1}{n^3 + 2}\).
- Highest powers: \(n\) on top, \(n^3\) below, so compare with \(b_n = \dfrac{n}{n^3} = \dfrac{1}{n^2}\).
- \(\dfrac{a_n}{b_n} = \dfrac{(2n + 1)\,n^2}{n^3 + 2} = \dfrac{2 + 1/n}{1 + 2/n^3} \to 2\), which is finite and non-zero.
- \(\sum \dfrac{1}{n^2}\) converges, so \(\sum a_n\) does too.
AnswerConverges.
Example 4: Ratio test with a power of 2
MediumTest \(\displaystyle\sum \frac{n^2}{2^n}\).
- \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)^2}{2^{n+1}}\cdot\dfrac{2^n}{n^2} = \dfrac{1}{2}\left(1 + \dfrac{1}{n}\right)^2 \to \dfrac{1}{2}\).
- \(l = \dfrac{1}{2} < 1\).
AnswerConverges, by the ratio test.
Example 5: Ratio test with factorials
Exam levelTest \(\displaystyle\sum \frac{n!}{n^n}\).
- \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)!}{(n + 1)^{n+1}}\cdot\dfrac{n^n}{n!} = \dfrac{(n + 1)\,n^n}{(n + 1)^{n+1}} = \left(\dfrac{n}{n + 1}\right)^n\).
- \(\left(\dfrac{n}{n + 1}\right)^n = \dfrac{1}{\left(1 + \frac{1}{n}\right)^n} \to \dfrac{1}{e}\).
- \(\dfrac{1}{e} \approx 0.37 < 1\).
AnswerConverges, by the ratio test.
Example 6: Root test
Exam levelTest \(\displaystyle\sum \left(\frac{n}{n + 1}\right)^{n^2}\).
- The whole term is a power of \(n\), so use the root test: \((a_n)^{1/n} = \left(\dfrac{n}{n + 1}\right)^{n}\).
- As in Example 5, this tends to \(\dfrac{1}{e} < 1\).
AnswerConverges, by the root test.
Example 7: An alternating series
MediumTest \(\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \cdots\)
- \(u_n = \dfrac{1}{n}\) is positive, decreasing, and tends to 0, so by Leibniz's test the series converges.
- But \(\sum |a_n| = \sum \dfrac{1}{n}\) is the harmonic series, which diverges.
AnswerConverges conditionally (not absolutely).
Choosing the test is half the marks. Ask in order: Do the terms tend to 0? Is it geometric or a \(p\)-series in disguise? Powers of \(n\) only? Use limit comparison. Factorials or \(a^n\)? Use the ratio test. The whole term raised to the power \(n\)? Use the root test. Alternating signs? Use Leibniz.
Common mistakes
1. Thinking \(a_n \to 0\) proves convergence
It's necessary but not enough. The harmonic series \(\sum \dfrac{1}{n}\) has \(a_n \to 0\) and still diverges.
2. Reading something into \(l = 1\)
When the ratio or root test gives \(l = 1\), it tells you nothing. \(\sum \dfrac{1}{n}\) and \(\sum \dfrac{1}{n^2}\) both give \(l = 1\); one diverges and the other converges. Switch to comparison or Raabe's test.
3. Using the ratio test on powers of \(n\)
For terms like \(\dfrac{2n + 1}{n^3 + 2}\), the ratio test always gives \(l = 1\). Limit comparison is the right tool.
4. Confusing the sequence and the series
The sequence \(\dfrac{n}{n + 1}\) converges (to 1), but the series \(\sum \dfrac{n}{n + 1}\) diverges.
How it's asked in exams
- Test the convergence of a given series, choosing the right test and stating its conclusion clearly.
- Find the values of \(x\) for which a series like \(\sum \dfrac{x^n}{n}\) converges.
- Absolute or conditional? Classify an alternating series.
Practice questions
Try each one on paper before you open the answer.
Q1Find the sum of \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n}\).
A geometric series with first term \(\tfrac{1}{2}\) and ratio \(\tfrac{1}{2}\). Sum \(= \dfrac{1/2}{1 - 1/2} = 1\).
Q2Test \(\displaystyle\sum \frac{1}{\sqrt n}\).
A \(p\)-series with \(p = \tfrac{1}{2} \le 1\). Diverges.
Q3Test \(\displaystyle\sum \frac{3^n}{n!}\).
Ratio: \(\dfrac{a_{n+1}}{a_n} = \dfrac{3}{n + 1} \to 0 < 1\). Converges.
Q4Test \(\displaystyle\sum \frac{n}{n^2 + 1}\).
Limit comparison with \(\dfrac{1}{n}\): \(\dfrac{n^2}{n^2 + 1} \to 1\). Since \(\sum \dfrac{1}{n}\) diverges, so does this series.
Q5Test \(\displaystyle\sum \left(1 + \frac{1}{n}\right)^{-n^2}\).
Root test: \((a_n)^{1/n} = \left(1 + \dfrac{1}{n}\right)^{-n} \to \dfrac{1}{e} < 1\). Converges.
Q6Is \(\displaystyle\sum \frac{(-1)^{n+1}}{\sqrt n}\) absolutely or conditionally convergent?
Leibniz: \(\dfrac{1}{\sqrt n}\) decreases to 0, so the series converges. But \(\sum \dfrac{1}{\sqrt n}\) diverges (\(p = \tfrac{1}{2}\)), so the convergence is conditional.
Q7Challenge. For which \(x > 0\) does \(\displaystyle\sum \frac{x^n}{n}\) converge?
Ratio: \(\dfrac{a_{n+1}}{a_n} = \dfrac{n}{n + 1}\,x \to x\). It converges for \(x < 1\) and diverges for \(x > 1\).
At \(x = 1\) the ratio test fails, but the series becomes the harmonic series, which diverges. So it converges exactly when \(0 < x < 1\).
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Frequently asked questions
What does it mean for an infinite series to converge?
Its partial sums \(S_n = a_1 + \cdots + a_n\) approach a finite limit. That limit is called the sum of the series.
Which test should I try first?
Always the \(n\)th-term test: if the terms don't tend to 0, you're done, because the series diverges. Then pick a test based on the form of \(a_n\), as in the tip in section 04.
Why does the harmonic series diverge if its terms go to zero?
The terms shrink too slowly. You can always group enough of them to add up to more than \(\tfrac{1}{2}\), as many times as you like.
What is the difference between absolute and conditional convergence?
Absolute: the series still converges when every term is made positive. Conditional: it converges only because positive and negative terms cancel.
Where series lead next
← All 33 Calculus topics
Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.