Grade 11

Limits of sequences

The value a sequence of numbers approaches as its index grows without bound.

IntuitionChasing a sequence to infinity

Take the sequence $a_n=\frac{1}{n}$ : $1,\ 0.5,\ 0.333\ldots,\ 0.25,\ldots$ The terms never actually become $0$ , yet they squeeze arbitrarily close to $0$ as $n$ grows. This "getting arbitrarily close forever" is exactly what the limit of a sequence captures, and it is the engine behind infinite sums, iterative algorithms, and compound interest.

Partial sums

S_n=\sum_{k=0}^{n-1}u_1q^k

of a geometric series; drag

n

and watch

S_n\to \frac{u_1}{1-q}

SchoolFormal definition

Definition: Limit of a sequence (ε–N definition)

We say $\lim_{n\to\infty} a_n = L$ if for every number $\varepsilon>0$ , no matter how small, there is an index $N$ beyond which every term $a_n$ differs from $L$ by less than $\varepsilon$ . Symbolically: $\forall \varepsilon>0,\ \exists N\in\mathbb{N}^*:\ \forall n>N,\ |a_n-L|<\varepsilon$

\forall \varepsilon>0,\ \exists N\in\mathbb{N}^*:\ \forall n>N,\ |a_n-L|<\varepsilon

Here $\varepsilon$ (epsilon) is an arbitrarily small tolerance chosen by a skeptic, $N$ is the "waiting index" we get to choose in response, and $|a_n-L|<\varepsilon$ says the term $a_n$ lies inside the strip $(L-\varepsilon, L+\varepsilon)$ . The definition works like a game: however small $\varepsilon$ the skeptic picks, we can always name an $N$ that works.

S=\lim_{n\to\infty}S_n=\frac{u_1}{1-q}

Special limits used constantly in practice
Form	Result
Power reciprocal	$\lim_{n\to\infty}\frac{1}{n^k}=0$ for $k>0$
Geometric decay	$\lim_{n\to\infty} q^n = 0$ when $\|q\|<1$
Constant sequence	$\lim_{n\to\infty} c = c$
Polynomial growth	$\lim_{n\to\infty} n^k = +\infty$ for $k>0$

UndergraduateTwo pillar theorems

Theorem: Monotone convergence theorem

If $(a_n)$ is increasing and bounded above, then it converges, and $a_1\le a_2\le\cdots\le M\ \Rightarrow\ \lim_{n\to\infty}a_n=\sup_n a_n$ (symmetrically for decreasing sequences bounded below).

Why is it true?

Intuitively, a sequence that only ever climbs but can never pass a ceiling has nowhere else to go except settle right below that ceiling.

Proof

Let $(a_n)$ be increasing and bounded above. By the completeness axiom of $\mathbb{R}$ , the set $\{a_n : n\in\mathbb{N}^*\}$ has a least upper bound $L=\sup_n a_n$ .

Fix $\varepsilon>0$ . Since $L$ is the least upper bound, $L-\varepsilon$ is not an upper bound, so there exists an index $N$ with $a_N>L-\varepsilon$ .

Because $(a_n)$ is increasing, for every $n>N$ we have $a_n\ge a_N>L-\varepsilon$ . Also $a_n\le L$ for every $n$ , since $L$ is an upper bound of the whole sequence.

Combining both bounds, $L-\varepsilon<a_n\le L<L+\varepsilon$ for all $n>N$ , i.e. $|a_n-L|<\varepsilon$ . Since $\varepsilon>0$ was arbitrary, $\lim_{n\to\infty} a_n = L$ holds by the ε–N definition. $\blacksquare$

Theorem: Squeeze theorem

If $b_n\le a_n\le c_n,\ \lim b_n=\lim c_n=L\ \Rightarrow\ \lim a_n=L$ , then $\lim_{n\to\infty} a_n = L$ as well.

Why is it true?

If a sequence is trapped between two other sequences that both converge to the same place, it has no room to go anywhere else.

Proof

Suppose $b_n\le a_n\le c_n$ for all sufficiently large $n$ , and $\lim b_n=\lim c_n=L$ .

Fix $\varepsilon>0$ . Since $\lim b_n=L$ , there is $N_1$ with $L-\varepsilon<b_n<L+\varepsilon$ for all $n>N_1$ . Since $\lim c_n=L$ , there is $N_2$ with $L-\varepsilon<c_n<L+\varepsilon$ for all $n>N_2$ .

Let $N=\max(N_1,N_2)$ . For every $n>N$ , combining $b_n\le a_n\le c_n$ with both inequalities above gives $L-\varepsilon<b_n\le a_n\le c_n<L+\varepsilon$ , hence $L-\varepsilon<a_n<L+\varepsilon$ , i.e. $|a_n-L|<\varepsilon$ .

Since $\varepsilon>0$ was arbitrary, $\lim_{n\to\infty} a_n=L$ by the ε–N definition. $\blacksquare$

UndergraduateReal-World Applications and Worked Examples

Sequence limits drive geometric series in finance (perpetuities, amortization), physics (energy lost in a bouncing ball), computer science (loop convergence, fixed-point iteration) and numerical analysis (root-finding algorithms that predate calculators, like the Babylonian method for square roots).

Example

A ball is dropped from height $h_0=2$ m. After each bounce it rises to $q=0.6$ times the height of the previous bounce ( $|q|<1$ ). Find the total distance the ball travels (falling and rising) before it comes to rest, assuming infinitely many bounces.

Solution

The first fall contributes $h_0$ . After that, each bounce $k=1,2,\ldots$ contributes an "up" trip of height $h_0q^k$ and a "down" trip of the same height, so it adds $2h_0q^k$ to the total.

Summing the infinitely many bounces is a geometric series: $\sum_{k=1}^{\infty} 2h_0q^k = 2h_0\cdot\frac{q}{1-q}$ , using $S=\lim_{n\to\infty}S_n=\frac{u_1}{1-q}$ with first term $q$ .

Total distance $=h_0+2h_0\cdot\frac{q}{1-q}=h_0\cdot\frac{1-q+2q}{1-q}=h_0\cdot\frac{1+q}{1-q}$ .

Substituting $h_0=2$ and $q=0.6$ : total $=2\cdot\frac{1.6}{0.4}=2\cdot 4=8$ m. Even though the ball bounces infinitely many times, it travels a finite total distance — a direct consequence of the geometric series limit.

Example

Consider the sequence defined by $u_1=1$ and $u_{n+1}=\frac{1}{2}\left(u_n+\frac{2}{u_n}\right)$ (the Babylonian / Heron algorithm for square roots). Show that $(u_n)$ converges and find its limit.

Solution

First, boundedness: by the AM–GM inequality, for any $u_n>0$ , $u_{n+1}=\frac{1}{2}\left(u_n+\frac{2}{u_n}\right)\ge\sqrt{u_n\cdot\frac{2}{u_n}}=\sqrt{2}$ , so $u_n\ge\sqrt{2}$ for every $n\ge 2$ .

Next, monotonicity: for $n\ge 2$ , $u_{n+1}-u_n=\frac{2-u_n^2}{2u_n}\le 0$ because $u_n^2\ge 2$ . So $(u_n)$ is decreasing (from index $2$ onward) and bounded below by $\sqrt{2}$ .

By the Monotone Convergence Theorem, $(u_n)$ converges to some limit $L$ . Taking the limit of both sides of the recurrence, $L=\frac{1}{2}\left(L+\frac{2}{L}\right)$ , which gives $2L=L+\frac{2}{L}$ , so $L^2=2$ , hence $L=\sqrt{2}$ (the positive root, since every $u_n>0$ ).

Numerically: $u_1=1,\ u_2=1.5,\ u_3\approx1.41667,\ u_4\approx1.414216$ — already accurate to five decimal places of $\sqrt{2}$ after just three steps, showing how fast this ancient iterative method converges.

Compute $\lim_{n\to\infty}\frac{2n+1}{n+3}$ .

For $q=0.5$ , what is $\lim_{n\to\infty} q^n$ ?

An infinite geometric series has first term $u_1=3$ and ratio $q=\frac{1}{3}$ . What is its sum?

Which extra condition, together with monotonicity, guarantees a sequence has a finite limit?

References

James Stewart (2015). Calculus: Early Transcendentals
Judith V. Grabiner (1983). Who Gave You the Epsilon? Cauchy and the Origins of Rigorous Calculus