Convergence of series; the nth-term test

About this lesson

Calculus BC (AP-aligned) › Infinite series and convergence tests

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Convergence of series; the nth-term test

Warm-up

Three quick questions from earlier lessons.

Discover

∑n=1∞nn+1=12+23+34+⋯\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{n + 1} = \tfrac{1}{2} + \tfrac{2}{3} + \tfrac{3}{4} + \dotsb. The terms get close to 1. Can the sum be a finite number?

  1. No
  2. Yes, since each term is less than 1

Key idea

A series converges when its partial sums settle.

∑an\displaystyle\sum a_n converges to SS if the partial sums SN=a1+⋯+aNS_N = a_1 + \dotsb + a_N tend to SS. nth-term test: if an↛0a_n \not\to 0, the series diverges. If an→0a_n \to 0, the test says nothing (∑ 1/n diverges).

Worked example

∑n=1∞(1n−1n+1)\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right), a telescoping sum.

SN=1−1N+1S_N = 1 - \dfrac{1}{N + 1}
Terms cancel in pairs.
SN→1S_N \to 1
N → ∞.
The series converges to 1
So:

Your turn, guided

Your turn: ∑3nn+2\displaystyle\sum \dfrac{3n}{n + 2}. The limit of the terms is…

33, so the series diverges
3n/(n + 2) → 3 ≠ 0: diverges.

Practice ★

  1. Does ∑n=1∞6nn+3\displaystyle\sum_{n=1}^{\infty} \dfrac{ 6n}{n + 3} converge?
    1. No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
    2. Yes, because each term is less than 6
    3. The nth-term test cannot decide
  2. Does ∑n=1∞5nn+8\displaystyle\sum_{n=1}^{\infty} \dfrac{ 5n}{n + 8} converge?
    1. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
    2. Yes, because each term is less than 5
    3. The nth-term test cannot decide
  1. Find ∑n=1∞(9n−9n+1)\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{ 9 }{n} - \dfrac{ 9 }{n + 1}\right).
  2. Find ∑n=1∞(3n−3n+1)\displaystyle\sum_{n=1}^{\infty} \left(\dfrac{ 3 }{n} - \dfrac{ 3 }{n + 1}\right).

Watch out!

Practice ★★

  1. ∑n=1∞4n\displaystyle\sum_{n=1}^{\infty} \dfrac{ 4 }{n}: its terms tend to 0. What does that tell you?
    1. Nothing by itself: this series in fact diverges (harmonic series)
    2. The series converges
    3. The sum is 0

Exit ticket

Three questions to check you have got it.

  1. Does ∑n=1∞6nn+3\displaystyle\sum_{n=1}^{\infty} \dfrac{ 6n}{n + 3} converge?
    1. No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
    2. Yes, because each term is less than 6
    3. The nth-term test cannot decide
  2. Does ∑n=1∞5nn+8\displaystyle\sum_{n=1}^{\infty} \dfrac{ 5n}{n + 8} converge?
    1. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
    2. Yes, because each term is less than 5
    3. The nth-term test cannot decide
  3. Does ∑n=1∞5nn+2\displaystyle\sum_{n=1}^{\infty} \dfrac{ 5n}{n + 2} converge?
    1. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
    2. Yes, because each term is less than 5
    3. The nth-term test cannot decide

Summary card

Convergence

  • Converges: the partial sums have a limit.
  • an↛0a_n \not\to 0: diverges. an→0a_n \to 0: test another way.

Watch out: the harmonic series diverges.

End of the lesson

Next lesson: The integral test and p-series Back to the chapter

Answers: Convergence of series; the nth-term test

A

  1. No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
  2. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)

B

  1. 99
  2. 33

C

  1. Nothing by itself: this series in fact diverges (harmonic series)

D

  1. No: the terms tend to 6 ≠ 0, so it diverges (nth-term test)
  2. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)
  3. No: the terms tend to 5 ≠ 0, so it diverges (nth-term test)

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