What a sequence is, and what convergence means
A sequence is an infinite list of numbers \(a_1, a_2, a_3, \dots\), usually given by a formula for the \(n\)th term \(a_n\). For example, \(a_n = \dfrac{1}{n}\) gives \(1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \dots\)
Zeno's old puzzle is a sequence. To cross a room you first walk half way, then half the remaining distance, and so on. The distances covered are \(\tfrac{1}{2}, \tfrac{3}{4}, \tfrac{7}{8}, \tfrac{15}{16}, \dots\), that is \(a_n = 1 - \dfrac{1}{2^n}\). The terms never reach 1, but they get as close to 1 as you like. That is convergence.
Convergent
A sequence \((a_n)\) converges to a limit \(L\), written \(\displaystyle\lim_{n \to \infty} a_n = L\), if for every \(\varepsilon > 0\) there is a number \(N\) such that \[|a_n - L| < \varepsilon \quad \text{for all } n > N\]
Divergent
A sequence that does not converge diverges. It may go to \(+\infty\) or \(-\infty\), or it may oscillate without settling, like \((-1)^n\).
In words: however small a band \(\varepsilon\) you draw around \(L\), the terms eventually enter the band and never leave it again.
The idea in one picture
Standard results and tools
Standard limits
| Sequence | Limit as \(n \to \infty\) |
|---|---|
| \(\dfrac{1}{n^p}\), with \(p > 0\) | \(0\) |
| \(r^n\) | \(0\) if \(|r| < 1\); \(1\) if \(r = 1\); diverges otherwise |
| \(a^{1/n}\), with \(a > 0\) | \(1\) |
| \(n^{1/n}\) | \(1\) |
| \(\left(1 + \dfrac{x}{n}\right)^n\) | \(e^x\) |
| \(\dfrac{\log n}{n}\) | \(0\) |
| \(\dfrac{n^k}{a^n}\), with \(a > 1\) | \(0\) |
| \(\dfrac{a^n}{n!}\) | \(0\) |
Bounded and monotonic sequences
- A sequence is bounded if there is a number \(M\) with \(|a_n| \le M\) for all \(n\).
- It is monotonic if it is always increasing (\(a_{n+1} \ge a_n\)) or always decreasing (\(a_{n+1} \le a_n\)).
- Every convergent sequence is bounded. The converse is false: \((-1)^n\) is bounded but does not converge.
A sequence that is monotonic and bounded always converges.
This theorem is how you handle sequences defined step by step (recursively), where there is no direct formula for \(a_n\). See Example 6.
Solved examples
Example 1: A rational expression
EasyFind \(\displaystyle\lim_{n \to \infty} \frac{2n + 3}{n + 1}\).
- Divide the top and bottom by the highest power of \(n\), which is \(n\): \(\dfrac{2 + 3/n}{1 + 1/n}\).
- As \(n \to \infty\), \(\dfrac{3}{n} \to 0\) and \(\dfrac{1}{n} \to 0\).
AnswerThe sequence converges to \(2\).
Example 2: Proving it from the definition
MediumUsing the \(\varepsilon\)–\(N\) definition, prove that \(\dfrac{2n + 3}{n + 1} \to 2\).
- \(\left|\dfrac{2n + 3}{n + 1} - 2\right| = \left|\dfrac{2n + 3 - 2n - 2}{n + 1}\right| = \dfrac{1}{n + 1}\).
- We need \(\dfrac{1}{n + 1} < \varepsilon\), that is \(n > \dfrac{1}{\varepsilon} - 1\).
- So choose \(N = \dfrac{1}{\varepsilon} - 1\) (or any larger whole number). For every \(n > N\), \(|a_n - 2| < \varepsilon\).
AnswerProved. For example, with \(\varepsilon = 0.01\), every term after \(n = 99\) is within 0.01 of 2.
Example 3: An oscillating sequence
EasyDoes \(a_n = (-1)^n\) converge?
- The terms are \(-1, 1, -1, 1, \dots\)
- They never settle near one value: any band of width less than 1 around a proposed limit misses infinitely many terms.
AnswerNo. It is bounded but oscillates, so it diverges.
Example 4: Rationalising
MediumFind \(\displaystyle\lim_{n \to \infty} \left(\sqrt{n + 1} - \sqrt{n}\right)\).
- This is \(\infty - \infty\), so multiply and divide by the conjugate: \(\dfrac{(n + 1) - n}{\sqrt{n + 1} + \sqrt n} = \dfrac{1}{\sqrt{n + 1} + \sqrt n}\).
- The denominator grows without bound.
AnswerThe sequence converges to \(0\).
Example 5: The number e
MediumFind \(\displaystyle\lim_{n \to \infty} \left(1 + \frac{2}{n}\right)^n\).
- Use the standard result \(\left(1 + \dfrac{x}{n}\right)^n \to e^x\), with \(x = 2\).
- Careful: the base tends to 1 and the power to \(\infty\). This “\(1^\infty\)” form is not automatically 1.
Answer\(e^2\)
Example 6: A recursive sequence
Exam levelLet \(a_1 = \sqrt 2\) and \(a_{n+1} = \sqrt{2 + a_n}\). Show that the sequence converges, and find its limit.
- Bounded above by 2 (by induction): \(a_1 = \sqrt 2 < 2\). If \(a_n < 2\), then \(a_{n+1} = \sqrt{2 + a_n} < \sqrt 4 = 2\).
- Increasing: \(a_{n+1}^2 - a_n^2 = 2 + a_n - a_n^2 = (2 - a_n)(1 + a_n) > 0\), since \(0 < a_n < 2\). So \(a_{n+1} > a_n\).
- Increasing and bounded, so by the monotone convergence theorem the sequence converges to some \(L\).
- Let \(n \to \infty\) in \(a_{n+1} = \sqrt{2 + a_n}\): \(L = \sqrt{2 + L}\), so \(L^2 - L - 2 = 0\), giving \(L = 2\) or \(L = -1\). All terms are positive, so \(L = 2\).
AnswerThe sequence converges to \(2\).
Common mistakes
1. Confusing a sequence with a series
A sequence is a list of terms \(a_n\). A series is their sum \(\sum a_n\). The sequence \(\dfrac{1}{n}\) converges to 0, but the series \(\sum \dfrac{1}{n}\) diverges. See convergence of series.
2. Thinking bounded means convergent
\((-1)^n\) is bounded but does not converge. You need bounded and monotonic for the guarantee.
3. Treating \(1^\infty\) as 1
\(\left(1 + \dfrac{1}{n}\right)^n\) tends to \(e\), not 1. Both the base and the power are moving, so this is an indeterminate form.
4. Assuming the limit exists before finding it
In Example 6 you may only solve \(L = \sqrt{2 + L}\) after proving the sequence converges. Otherwise the “limit” you find may not exist.
How it's asked in exams
- Test for convergence and find the limit of a given sequence.
- Prove a limit using the \(\varepsilon\)–\(N\) definition.
- Show a sequence is monotonic and bounded, often a recursive one, and find its limit.
- Classify a sequence as convergent, divergent or oscillating.
Practice questions
Try each one on paper before you open the answer.
Q1Find \(\displaystyle\lim_{n \to \infty} \frac{3n^2 - n}{5n^2 + 2}\).
Divide by \(n^2\): \(\dfrac{3 - 1/n}{5 + 2/n^2} \to \dfrac{3}{5}\).
Q2Find \(\displaystyle\lim_{n \to \infty} \frac{n}{n^2 + 1}\).
Divide by \(n^2\): \(\dfrac{1/n}{1 + 1/n^2} \to 0\).
Q3Does \(a_n = \dfrac{(-1)^n}{n}\) converge?
Yes, to 0. The sign alternates, but \(|a_n| = \dfrac{1}{n} \to 0\).
Q4Find \(\displaystyle\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n\).
Use \(\left(1 + \dfrac{x}{n}\right)^n \to e^x\) with \(x = -1\): the limit is \(\dfrac{1}{e}\).
Q5Find \(\displaystyle\lim_{n \to \infty} n \sin\frac{\pi}{n}\).
Write it as \(\pi \cdot \dfrac{\sin(\pi/n)}{\pi/n}\). Since \(\dfrac{\sin t}{t} \to 1\) as \(t \to 0\), the limit is \(\pi\).
Q6Show that \(a_n = \dfrac{n}{n + 1}\) is increasing and bounded, and find its limit.
\(a_{n+1} - a_n = \dfrac{n + 1}{n + 2} - \dfrac{n}{n + 1} = \dfrac{1}{(n + 1)(n + 2)} > 0\), so it is increasing. Also \(a_n < 1\) for all \(n\).
So it converges, and \(\dfrac{n}{n + 1} = \dfrac{1}{1 + 1/n} \to 1\).
Q7Challenge. Show that \(\dfrac{2^n}{n!} \to 0\).
For \(n \ge 2\), \(\dfrac{a_{n+1}}{a_n} = \dfrac{2}{n + 1} \le \dfrac{2}{3}\). So each term is at most \(\tfrac{2}{3}\) of the one before, and \(a_n \le a_2\left(\tfrac{2}{3}\right)^{n-2} \to 0\). Since every term is positive, \(a_n \to 0\).
Bring it to a free demo class and work through it with Vipul Sir. Book on WhatsApp.
Frequently asked questions
What does it mean for a sequence to converge?
Its terms get closer and closer to one fixed number, the limit, and eventually stay as close to it as you like.
What is the difference between a sequence and a series?
A sequence is a list of numbers. A series is the sum of the terms of a sequence. They converge under different conditions.
Can a sequence have two different limits?
No. If a sequence converges, its limit is unique.
Is every bounded sequence convergent?
No. \((-1)^n\) is bounded but oscillates. A bounded sequence that is also monotonic does converge.
Where sequences 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.