Modules › Sequences and infinite series
Sequences and infinite series
Limits of sequences, series, geometric and telescoping series, and the convergence tests.
Lectures 10b–11 · Tutorial 11 · Sequences, series, power series and Taylor series
Textbook sections: Thomas ch. 10 Infinite Sequences and Series (10.1–10.10) · Stewart ch. 11 Infinite Sequences and Series (11.1–11.11)
Textbook sections: Thomas ch. 10 Infinite Sequences and Series (10.1–10.10) · Stewart ch. 11 Infinite Sequences and Series (11.1–11.11)
Sequences and infinite series
Calculus I · Module 12
After this module you can
- Find limits of explicit and recursive sequences.
- Sum geometric and telescoping series.
- Choose the right convergence test (n-th term, integral, comparison, limit comparison, ratio, root, alternating) and justify it.
Sequences
- $a_n\to L$ if $|a_n-L|<\varepsilon$ for all large $n$. Standard: $\sqrt[n]{n}\to1$, $r^n\to0$ for $|r|<1$, $\left(1+\frac an\right)^n\to e^a$, $\frac{n!}{n^n}\to0$.
- Squeeze: $\frac{\sin n}{n}\to0$. Exponentials beat polynomials: $\frac{3^n}{n^3}\to\infty$.
- Recursive $a_{n+1}=f(a_n)$ (assuming convergence): solve $L=f(L)$ and discard roots incompatible with the signs of the terms: $a_{n+1}=\frac{a_n+6}{a_n+2}\to2$ (not $-3$).
- Monotone and bounded $\Rightarrow$ convergent.
Series: sums
- $\sum a_n$ converges if the partial sums $S_N$ converge. Necessary: $a_n\to0$ (the $n$-th term test proves divergence only).
- Geometric: $\sum_{n\ge0}ar^n=\frac{a}{1-r}$ for $|r|<1$. Example: $0.\overline{23}=\frac{23}{99}$.
- Telescoping: $\sum\frac{5}{n(n+1)}=5\sum\left(\frac1n-\frac1{n+1}\right)=5$.
- Harmonic series $\sum\frac1n$ diverges; $p$-series converges iff $p>1$.
Which test?
| Looks like | Use |
|---|---|
| $a_n\not\to0$ | $n$-th term test: diverges |
| $p$-series, $\frac{1}{n^p}$-like | comparison / limit comparison |
| $f(n)$ with $f$ positive decreasing, easy $\int f$ | integral test |
| factorials, $c^n$ | ratio test |
| $n$-th powers | root test |
| $(-1)^nb_n$ | alternating series test ($b_n\downarrow0$) |
Worked verdicts
- $\sum\frac{n!}{10^n}$: ratio $\frac{n+1}{10}\to\infty$: diverges. $\sum\frac{n^n}{(2^n)^2}$: root $\frac n4\to\infty$: diverges.
- $\sum ne^{-n^2}$: integral $\int_1^\infty xe^{-x^2}dx=\frac1{2e}$: converges. $\sum\frac{\cos^2n}{n^{3/2}}$: comparison with $n^{-3/2}$.
- $\sum\frac{1}{\sqrt n\ln n}$: $\ge\frac1{n^{3/4}}$ eventually: diverges. $\sum\frac{n^{\sqrt2}}{2^n}$: ratio $\to\frac12$: converges.
- $\sum\frac{1}{2^{\ln n}}=\sum n^{-\ln2}$: $p=\ln2<1$: diverges.
Typical mistakes
$a_n\to0$Does NOT imply convergence (harmonic series).
Ratio testInconclusive when the ratio tends to $1$.
Recursive limitsBoth roots of $L=f(L)$ may appear: use signs and monotonicity.
Practice next
- Chapter quizzes Ch. 6 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Series: partial sums and test hints
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Limits of sequences (Tutorial 11, Q2)
Let $a_1=-1$, $a_{n+1}=\dfrac{a_n+6}{a_n+2}$ and assume $(a_n)$ converges. Its limit is
Show the answer and the solution
Answer: $2$
If $a_n\to L$ then $L=\frac{L+6}{L+2}\Rightarrow L^2+L-6=0\Rightarrow L=2$ or $L=-3$. The terms $-1,5,\frac{11}7,\dots$ are positive from $a_2$ on, so the limit must be the positive root $2$.
Worked example 2 · Convergence tests (Computation problems 11, Q11)
Let $a_1=1$ and $a_{n+1}=\dfrac{1+\tan^{-1}n}{n}\,a_n$ ($n\ge1$). Does $\sum a_n$ converge?
Show the answer and the solution
Answer: converges (ratio test: $\frac{a_{n+1}}{a_n}\to0<1$)
$\frac{a_{n+1}}{a_n}=\frac{1+\arctan n}{n}\le\frac{1+\pi/2}{n}\to0$. So $\sum a_n$ converges (the partial sums stabilise: $a_2\approx1.785$, $a_3\approx1.88$, $a_4\approx1.41$, $a_5\approx0.82,\dots$).
Worked example 3 · Series sums (Final tutorial (session 2), differentiating a geometric series)
Differentiating $\dfrac{1}{1-x}=\sum_{n\ge0}x^n$ gives $\dfrac{1}{(1-x)^2}=\sum_{n\ge1}nx^{n-1}$ for $|x|<1$. Use this to find $\displaystyle\sum_{n=1}^\infty\dfrac{n}{2^n}$.
Show the answer and the solution
Answer: $2$
Multiply by $x$: $\sum_{n\ge1}nx^n=\dfrac{x}{(1-x)^2}$. At $x=\frac12$: $\dfrac{1/2}{(1/2)^2}=2$. (Forgetting the factor $x$ gives $\sum nx^{n-1}=4$.)
Worked example 4 · Limits of sequences (Computation problems 11, Q3)
The sequence $2,\ 2+\frac12,\ 2+\frac{1}{2+\frac12},\ 2+\frac{1}{2+\frac{1}{2+\frac12}},\dots$ (so $a_{n+1}=2+\frac1{a_n}$) converges to
Show the answer and the solution
Answer: $1+\sqrt2$
If $a_n\to L$: $L=2+\frac1L\Rightarrow L^2-2L-1=0\Rightarrow L=1\pm\sqrt2$; all terms are positive, so $L=1+\sqrt2$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Series: partial sums and test hints