Unit 11 · Term 2

Sequences

Sequences bring the ε–δ idea back, now as ε–N: a tail that stays arbitrarily close to a limit. They are the foundation for series in the final units.

By the end you can

State the ε–N definition of a sequence limit.
Use the limit laws and squeeze theorem for sequences.
Apply the Monotone Convergence Theorem.
Handle recursively defined sequences.

Why it matters & where it comes from

Why it exists

Sequences make "getting arbitrarily close" precise in the simplest setting, and they are the scaffolding for series. Cauchy sequences are even used to build the real numbers themselves.

In the real world

Every iterative algorithm is a sequence converging (you hope) to an answer — Newton's method, fixed-point iteration, and the training loops of machine learning all live here. The convergence rate is exactly how fast they finish.

Story & open questions

Zeno's paradox (~450 BC) — Achilles can never catch the tortoise — is really a question about the limit of a sequence, and the limit resolves it. Two millennia later, Cantor and Dedekind used Cauchy sequences to construct $\mathbb{R}$ rigorously. The ratio of consecutive Fibonacci numbers tends to the golden ratio, a limit that also appears in mathematical models of phyllotaxis.

How sequence proofs work

A sequence limit proof is an $\varepsilon$ proof with only one moving variable: the index $n$. The proof should show that every term far enough out in the tail is close to the proposed limit.

Epsilon-N routine

Start from $|a_n-L|<\varepsilon$ and solve for a lower bound on $n$.

Choose a natural number $N$ larger than that bound.

Prove that every $n\ge N$ satisfies the desired inequality.

Recursive sequences

For recurrences, finding a candidate limit is not enough. First prove the sequence actually converges, usually by monotone convergence or by contraction estimates. Only then is it legitimate to take limits on both sides of the recurrence.

Find a stable interval: prove all terms stay inside it.
Prove monotonicity or another convergence mechanism.
Let the limit be $L$ and solve the fixed-point equation.
Reject algebraic roots that do not fit the interval or sign of the sequence.

The ε–N definition

Definition Sequence limit

$\lim_{n\to\infty}a_n=L$ means: for every $\varepsilon>0$ there is $N\in\mathbb N$ such that $n\ge N\Rightarrow|a_n-L|<\varepsilon$.

Example

Prove $\lim\dfrac{n}{n+1}=1$. Given $\varepsilon>0$, $\left|\dfrac{n}{n+1}-1 \right|=\dfrac{1}{n+1}<\varepsilon$ once $n>\tfrac1\varepsilon-1$, so take $N=\lceil1/\varepsilon\rceil$.

Limit laws and squeeze

Tools

Limits of sums, products, and quotients behave as expected. Squeeze: if $b_n\le a_n\le c_n$ and $b_n,c_n\to L$, then $a_n\to L$.

Example

$\dfrac{\sin n}{n}\to0$ since $-\dfrac1n\le\dfrac{\sin n}{n}\le\dfrac1n$.

Monotone Convergence Theorem

Theorem MCT

A sequence that is increasing and bounded above converges (to its supremum). Likewise decreasing and bounded below.

Example A recursive sequence

$a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.

By induction $a_n<2$ (bounded) and $a_{n+1}>a_n$ (increasing), so by MCT it converges to some $L$.

Taking limits in the recursion, $L=\sqrt{2+L}$, so $L^2-L-2=0$, giving $L=2$.

Common pitfalls

• $N$ may depend on $\varepsilon$ but not on $n$.
• A bounded sequence need not converge (e.g. $(-1)^n$); monotonicity is what MCT adds.
• When solving $L=f(L)$, also confirm the limit exists first.

Practice

Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.

Practice 1 · Definition

Prove $\lim\dfrac{2n+1}{n}=2$ from the definition.

Solution

$\left|\dfrac{2n+1}{n}-2\right|=\dfrac1n<\varepsilon$ for $n>1/\varepsilon$.

Practice 2 · Squeeze

Does $a_n=\dfrac{n!}{n^n}$ converge? To what?

Hint

$0<a_n\le\dfrac1n$ (each factor $\le1$, one equals $1/n$), so by squeeze $a_n\to0$.

Practice 3 · Recurrence

$a_1=1$, $a_{n+1}=\tfrac12(a_n+2/a_n)$. Show it converges and find the limit.

Answer

All terms are positive, and $a_2=\tfrac32>\sqrt2$. AM-GM gives $a_{n+1}\ge\sqrt2$, and whenever $a_n\ge\sqrt2$, $a_{n+1}\le a_n$. Thus the tail is decreasing and bounded below, so it converges. Its positive limit solves $L=\tfrac12(L+2/L)$, giving $L=\sqrt2$.

Practice 4 · Definition

Prove $\lim\dfrac{n^2}{n^2+1}=1$ directly from the definition.

Solution

$\left|\dfrac{n^2}{n^2+1}-1\right|=\dfrac{1}{n^2+1}<\dfrac{1}{n}<\varepsilon$ once $n>1/\varepsilon$.

Practice 5 · Growth rates

Evaluate $\displaystyle\lim_{n\to\infty}\frac{n!+2e^n}{3n!+4e^n}$.

Answer

Divide by $n!$. The ratio $\dfrac{e^n}{n!}\to0$ (its successive ratios are $\dfrac{e}{n+1}\to0$), so the limit is $\dfrac{1+0}{3+0}=\dfrac13.$

Practice 6 · Growth rates

Evaluate $\displaystyle\lim_{n\to\infty}\frac{5n^5+5^n+5\,n!}{n^n}$.

Answer

$n^n$ grows faster than each numerator term: $\dfrac{5n^5}{n^n}\to0$, $\dfrac{5^n}{n^n}=\left(\dfrac5n\right)^n\to0$, and $\dfrac{n!}{n^n}=\dfrac1n\cdot\dfrac2n\cdots\dfrac nn\le \left(\tfrac12\right)^{(n-1)/2}\to0$. Hence the limit is $0$.

Practice 7 · Theory

Prove: if a sequence is increasing and unbounded above, it diverges to $\infty$.

Solution

Given $M$, unboundedness gives some $a_N>M$; since the sequence increases, $a_n\ge a_N>M$ for all $n\ge N$. That is the definition of $a_n\to\infty$.

Practice 8 · Recurrence

$R_0=1$, $R_{n+1}=\dfrac{R_n+2}{R_n+3}$. Show it converges and find the limit.

Solution

Let $L=\sqrt3-1$, the positive fixed point of $g(x)=\dfrac{x+2}{x+3}$. For $x>L$, one checks that $L<g(x)<x$. Since $R_0=1>L$, induction shows that $(R_n)$ decreases and remains above $L$, so it converges. Taking limits in the recurrence gives $R_n\to L=\sqrt3-1$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:

Sequence limits. Use epsilon definitions and limit laws for sequences.
Monotone convergence. Know when monotone bounded sequences converge and how to find the limit.
Cauchy viewpoint. Use Cauchy behavior as a convergence criterion in complete settings.
Recursive sequences. Analyze fixed points, monotonicity, and boundedness before passing limits through recurrences.

Textbook/source alignment

Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.

Core theorem map

The sequence spine is: define convergence precisely, use monotone bounded arguments for existence, then identify the only possible limit by algebra.

Study routine. For recursive sequences, never pass to the limit before proving the sequence actually converges.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Epsilon proof

Prove $1/n\to0$.

Guide

Given $\epsilon>0$, choose $N>1/\epsilon$. If $n\ge N$, then $|1/n-0|\le1/N<\epsilon$.

Extension 2 · Monotone bounded

Show $a_n=1-1/n$ converges using monotone bounded convergence.

Guide

The sequence increases and is bounded above by $1$, so it converges. Directly, the limit is $1$.

Extension 3 · Cauchy

Explain why every convergent sequence is Cauchy.

Guide

If $a_n\to L$, make $|a_n-L|<\epsilon/2$ and $|a_m-L|<\epsilon/2$ for large $m,n$; then use the triangle inequality.

Extension 4 · Recursive caution

If $a_{n+1}=\sqrt{2+a_n}$ and $a_1=1$, what must be proved before solving $L=\sqrt{2+L}$?

Guide

You must prove convergence, usually by showing the sequence is increasing and bounded above by $2$. Then the fixed-point equation gives $L=2$.

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