Limit Theorems for Sequences

In this lesson you’ll learn the algebraic limit theorems, how one of them is proved from the definition, the order limit theorem, and the squeeze theorem. Together these let you compute almost every limit you’ll meet without returning to epsilon-N.

Proving $\frac{3n^2+2n}{n^2+1} \to 3$ straight from the definition is unpleasant. Proving it from the algebraic limit theorems takes two lines.

The point of this lesson is leverage. You pay the epsilon-N cost once, proving a handful of general results, and then compute freely.

Suppose $a_n \to A$ and $b_n \to B$. Then

$$a_n + b_n \to A + B$$ $$c \cdot a_n \to cA \quad \text{for any constant } c$$ $$a_n b_n \to AB$$ $$\frac{a_n}{b_n} \to \frac{A}{B} \quad \text{provided } B \ne 0$$

The quotient rule needs $B \ne 0$, and it also needs $b_n \ne 0$ eventually - which is automatic, since if $b_n \to B \ne 0$ then the terms are eventually within $\frac{|B|}{2}$ of $B$ and so nonzero.

These require both limits to exist. $a_n = (-1)^n$ and $b_n = (-1)^{n+1}$ both diverge, yet $a_n + b_n = 0$ converges. So you cannot run the theorem backwards.

Proof. Let $\varepsilon>0$. Since $a_n \to A$, there is $N_1$ with $|a_n - A| < \frac{\varepsilon}{2}$ for $n > N_1$. Since $b_n \to B$, there is $N_2$ with $|b_n - B| < \frac{\varepsilon}{2}$ for $n > N_2$.

Let $N = \max(N_1, N_2)$. For $n > N$, the triangle inequality gives

$$|(a_n + b_n) - (A+B)| = |(a_n - A) + (b_n - B)| \le |a_n - A| + |b_n - B| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$

∎

That’s why you split $\varepsilon$ in half: each piece is budgeted for one source of error, and they add to exactly $\varepsilon$. Applying the definition with $\frac{\varepsilon}{2}$ instead of $\varepsilon$ is legal because the definition holds for every positive tolerance.

The product needs one extra idea. Write

$$|a_nb_n - AB| = |a_nb_n - a_nB + a_nB - AB| \le |a_n||b_n - B| + |B||a_n - A|$$

Adding and subtracting $a_nB$ is the same bridging trick as before. The wrinkle is the factor $|a_n|$, which is not constant. But $(a_n)$ converges, so by the previous lesson it is bounded, say $|a_n| \le M$. Then

$$|a_nb_n - AB| \le M|b_n - B| + |B||a_n - A|$$

and each term can be forced below $\frac{\varepsilon}{2}$. Notice that “convergent implies bounded” was not a throwaway result; it is exactly what makes the product rule work.

If $a_n \to A$ and $a_n \ge 0$ for all $n$, then $A \ge 0$.

More generally, if $a_n \le b_n$ for all $n$ and both converge, then $\lim a_n \le \lim b_n$.

Weak inequalities survive limits; strict ones do not. Take $a_n = 0$ and $b_n = \frac1n$. Then $a_n < b_n$ strictly for every $n$, yet both limits are 0, so the conclusion is $0 \le 0$ and not $0 < 0$.

This trips people up constantly. Passing to a limit can turn $<$ into $=$, never into $>$.

If $b_n \le a_n \le c_n$ for all $n$, and $b_n \to L$ and $c_n \to L$, then $a_n \to L$.

This is the most useful tool in the lesson, because it needs no information about $a_n$ beyond the sandwich. You don’t have to understand $a_n$ at all; you only have to trap it.

Proof sketch. Given $\varepsilon>0$, get $N_1$ past which $b_n > L - \varepsilon$ and $N_2$ past which $c_n < L + \varepsilon$. Beyond $\max(N_1,N_2)$,

$$L - \varepsilon < b_n \le a_n \le c_n < L + \varepsilon$$

so $|a_n - L| < \varepsilon$. ∎

The standard application is anything with a bounded oscillating factor: $\left|\frac{\sin n}{n}\right| \le \frac1n$, so $-\frac1n \le \frac{\sin n}{n} \le \frac1n$ and the middle goes to 0. There is no formula for $\sin n$ as $n$ runs over the integers, and it doesn’t matter.

If $a_n \to 0$ and $(b_n)$ is bounded, then $a_nb_n \to 0$.

Note $(b_n)$ need not converge. $\frac{(-1)^n}{n} \to 0$ even though $(-1)^n$ diverges, because $\frac1n \to 0$ and the other factor stays bounded. The product rule doesn’t apply here; this result does.

Example 1: Compute $\lim \frac{3n^2+2n}{n^2+1}$.

Solution. Divide numerator and denominator by the highest power, $n^2$:

$$\frac{3n^2+2n}{n^2+1} = \frac{3 + \frac{2}{n}}{1 + \frac{1}{n^2}}$$

Now $\frac2n \to 0$ and $\frac{1}{n^2} \to 0$, so by the sum rule the numerator $\to 3$ and the denominator $\to 1$. Since the denominator’s limit is nonzero, the quotient rule gives

$$\lim \frac{3n^2+2n}{n^2+1} = \frac{3}{1} = 3$$

Compare Example 3 of the previous lesson, which proved a similar statement from the definition. Same result, a fraction of the work.

Example 2: Compute $\lim \frac{\sin n}{n}$.

Solution. The quotient rule is unavailable, since $\sin n$ has no limit as $n$ runs over the integers.

Instead squeeze. Since $|\sin n| \le 1$,

$$-\frac1n \le \frac{\sin n}{n} \le \frac1n$$

Both outer sequences $\to 0$, so the limit is 0. ∎

Example 3: Compute $\lim \frac{n!}{n^n}$.

Solution. Write out the ratio:

$$\frac{n!}{n^n} = \frac{n(n-1)(n-2)\cdots 1}{n \cdot n \cdots n} = 1 \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \frac1n$$

Every factor is at most 1, and the last is exactly $\frac1n$. So

$$0 < \frac{n!}{n^n} \le \frac1n$$

Squeeze gives 0. ∎

An estimate plus a squeeze beats trying to compute directly.

Example 4: A limit needing care with the quotient rule.

Compute $\lim \frac{n^2+1}{2n+3}$.

Solution. Dividing by $n$:

$$\frac{n^2+1}{2n+3} = \frac{n + \frac1n}{2 + \frac3n}$$

The denominator $\to 2$, but the numerator diverges to infinity, so the quotient rule does not apply - it requires both limits to exist.

The sequence diverges to $\infty$. To see it directly, for $n \ge 3$,

$$\frac{n^2+1}{2n+3} > \frac{n^2}{2n+3n} = \frac{n}{5}$$

which is unbounded. Divergent. ∎

The lesson: check the hypotheses. A theorem misapplied gives a confident wrong answer.

Example 5: Strict inequality lost in the limit.

Both $a_n = \frac{1}{n+1}$ and $b_n = \frac1n$ are positive with $a_n < b_n$ for every $n$. What does the order limit theorem give?

Solution. Both converge to 0, so the conclusion is $0 \le 0$. True but not strict.

Strict inequalities are not preserved. Any argument concluding $\lim a_n < \lim b_n$ from $a_n < b_n$ is wrong. To get a strict conclusion you need a gap that does not shrink, such as $a_n + c \le b_n$ for a fixed $c > 0$.

Example 6: A bounded times a null sequence.

Compute $\lim \frac{(-1)^n n}{n^2+1}$.

Solution. Rewrite as a product:

$$\frac{(-1)^n n}{n^2+1} = (-1)^n \cdot \frac{n}{n^2+1}$$

The second factor is $\frac{1/n}{1 + 1/n^2} \to 0$. The first is bounded by 1 though divergent.

By the bounded-times-null result, the limit is 0. ∎

Alternatively squeeze: $\left|\frac{(-1)^nn}{n^2+1}\right| \le \frac{n}{n^2} = \frac1n \to 0$.

Example 7: Why divergent sequences can sum to a convergent one.

Solution. $a_n = n$ and $b_n = -n$ both diverge, but $a_n + b_n = 0$ for every $n$, which converges to 0.

So the sum rule is a one-way street: convergence of both parts gives convergence of the sum, but convergence of the sum says nothing about the parts. Writing $\lim(a_n+b_n) = \lim a_n + \lim b_n$ is only meaningful once both right-hand limits are known to exist.

Asymptotic analysis of algorithms. Big-O statements are limit computations. Showing $\frac{3n^2+2n}{n^2} \to 3$ is the formal content of “this algorithm is $\Theta(n^2)$,” and dividing by the dominant term is the standard technique in both settings.

Numerical error analysis. Total error is usually a sum of a truncation term and a rounding term, and bounding each below $\frac{\varepsilon}{2}$ is exactly the sum-rule proof. That is why error bounds in numerical texts so often appear as a sum of two halves.

Signal processing. A bounded signal multiplied by a decaying envelope goes to zero regardless of the signal’s own behaviour, which is the bounded-times-null result and the reason exponential windowing damps arbitrary noise.

Physical estimates and sandwiching. Engineers bound an unknown quantity between two computable ones and let both converge. The squeeze theorem is the formal justification, and it is why “bracket the answer” is a legitimate method rather than a fudge.

Statistics. Consistency proofs for estimators typically squeeze the estimation error between two vanishing bounds. The estimator’s own formula may be intractable; only the sandwich matters.