Basic Analysis: Sequence Convergence (3)

So far, we’ve been considering the case where a sequence converges to a real number L. It’s also possible for a sequence to approach +∞ or -∞. The infinity symbol “∞” should be thought of as a convenient symbol instead of an actual infinity; in particular, it bears no relation to the cardinality (i.e. size) of the set of integers.

Definition. Let (a_n) be a sequence of real numbers. We write (a_n) → ∞ if

for every real M, there exists N such that if n>N, then a_n>M.

By the same token, (a_n) → -∞ if

for every real M, there exists N such that if n>N, then a_n<M.

[ A sequence which approaches +∞. ]

Let’s define a shorthand here: we say that “X eventually happens” for a sequence (a_n) if there exists N such that for all n > N, X happens for a_n. It should be apparent that if “X eventually happens” and “Y eventually happens” are both true, then “(X and Y) eventually happen” is also true.

Warning: if we have infinitely many statements of the form “for every m, X_m eventually happens”, one should not write “X_m eventually happens for all m” due to ambiguity in the English language. The latter statement can also be interpreted as “eventually, we have (X₁ and X₂ and X₃ …)”.

To illustrate the difference, consider the definition of (a_n) → ∞ which says: for every m, eventually a_n>m. This is different from saying: we eventually have (a_n>m for all m) since under this interpretation, no sequence of real numbers can satisfy it.

Now the three different types of limits can be summarised as follows:

We write (a_n) → $\left\{\begin{matrix} L\\+\infty\\-\infty\end{matrix}\right\}$ if for every $\left\{\begin{matrix} \epsilon>0\\ M \\ M\end{matrix}\right\}$ we eventually have $\left\{\begin{matrix}|a_n - L|<\epsilon.\\ a_n > M.\\ a_n<M.\end{matrix}\right\}$

Properties.

The limits satisfy the following properties, as expected from intuition.

If (a_n) → ∞ or L, and (b_n) → ∞, then (a_n+b_n) → ∞.
If (a_n) → ∞ or L>0, and (b_n) → ∞, then (a_nb_n) → ∞.
If (a_n) → -∞ or L<0, and (b_n) → ∞, then (a_nb_n) → -∞.
If (a_n) → ∞ or -∞, then (1/a_n) → 0.
If (a_n) → 0 for a sequence of positive numbers (a_n), then (1/a_n) → ∞.

Proof (Sketch).

To prove the first property, suppose (a_n) → L;

we eventually have |a_n–L| < 1, so a_n>L-1;
for any M, we eventually have b_n>M-(L-1).

Hence, for any M we eventually have a_n+b_n> M which proves that (a_n+b_n) → ∞. For the case (a_n) → ∞, replace the statement at the first bullet with “eventually, a_n>0″.

For the second property: suppose (a_n) → L>0. Eventually, |a_n–L| < L/2, so a_n>L/2, and for any M>0, eventually b_n>2M/L. Hence, for any M, eventually a_nb_n> M. This proves that (a_nb_n) → ∞.

The third property is similar to the second. Since (a_n) → L<0, eventually we have |a_n–L| < –L/2, so a_n<L/2. For any M<0, we also eventually have b_n>2M/L. Thus a_nb_n< M eventually.

For the fourth property: suppose (a_n) → ∞ or -∞. In the first case, for any M>0, we eventually have a_n>M>0, which gives 0 < 1/a_n<1/M. Thus, for any ε>0, we eventually have 0 < 1/a_n< ε. This proves that (a_n) → 0.

We’ll leave the remaining cases to the reader. ♦

Corollaries

If (a_n) → ∞ (resp. -∞), then (-a_n) → -∞ (resp. ∞). [ Multiply a_n by the constant sequence of (-1). ]
If (a_n) → ∞ or L, and (b_n) → -∞, then (a_n–b_n) → ∞. [ Write a_n–b_n= a_n+(-b_n) and apply two properties. ]

More Properties.

Correspondingly, we now have:

(Monotone convergence) An unbounded increasing sequence (a_n) must approach ∞. [In conjunction with the earlier result, an increasing sequence must have a limit L or ∞.]
(Convergence implies bounded) If (a_n) → ∞, then (a_n) has a lower bound.
(Squeeze theorem) If (a_n) and (b_n) are two sequences such that a_n ≥ b_n and (b_n) → ∞, then (a_n) → ∞.

Proof.

Pick M. Since (a_n) is unbounded, there exists N such that a_N > M. Hence, eventually we have a_n ≥ a_N > M. So (a_n) → ∞ as expected.
Eventually we have a_n>0. Since only finitely many terms are left out, (a_n) has a lower bound.
For any M, eventually we have b_n ≥ M. This implies a_n ≥ b_n ≥ M. ♦

We’ll leave it to the reader to write down the corresponding cases for (a_n) → -∞.

Limits Inferior and Superior

Let (a_n) be a sequence of real numbers.

Definition. Define a new sequence $b_n=\sup\{a_n, a_{n+1}, a_{n+2},\ldots\}$ . Clearly $b_1 \ge b_2 \ge b_3 \ge \ldots$ since we’re taking the sup of fewer elements as we go along. Since (b_n) is a decreasing sequence, it must have a limit. The limit superior of (a_n), written $\lim \sup (a_n)$ , is defined by $\lim (b_n)$ .

By the same token, let $c_n=\inf\{a_n, a_{n+1}, a_{n+2},\ldots\}$ . Now $c_1 \le c_2 \le c_3 \le\ldots$ so it has a limit, which is the limit inferior of (a_n), written $\lim\inf (a_n)$ .

Here’s a pictorial representation of lim sup and lim inf.

Let’s look at lim sup and lim inf in further detail.

First suppose (a_n) isn’t upper-bounded. Now even if we drop finitely many terms, the resulting sequence still isn’t upper-bounded. This gives $b_1 = b_2 = \ldots = +\infty$ . In this case, the limit superior is +∞.

If (a_n) is upper-bounded, then each b_n is a real number and we get a decreasing sequence (b_n). The limit of this sequence may be finite or -∞. If it’s -∞, for any M, eventually b_n < M and hence a_n ≤ b_n< M. So the lim sup = -∞ only occurs if a_n→ -∞.

In particular, observe that the lim sup is well-defined for all sequences.

In summary, we have the following cases:

Case 1 – (a_n) is not upper-bounded: lim sup (a_n) = +∞.
Case 2 – (a_n) → -∞: lim sup (a_n) = -∞.
Case 3 – otherwise: lim sup (a_n) = L, a real value.

[ Warning: even if (a_n) is not lower-bounded, it’s possible for its lim sup to be finite. See the third example later. ]

By the same token, for the limit inferior of (a_n) we get three cases.

Case 1 – (a_n) is not lower-bounded: each (c_n) = -∞, so lim inf (a_n) = -∞.
Case 2 – (a_n) → +∞: lim inf (a_n) = +∞.
Case 3 – otherwise: lim inf (a_n) = L, a real value.

Examples

Suppose $a_1 \le a_2 \le a_3 \le\ldots$ is an increasing sequence which converges to a finite L.Suppose a_n= (-1)ⁿ, i.e. the sequence is -1, +1, -1, +1, … . Then we have: for all n, b_n = +1 and c_n = -1. Thus, lim sup (a_n) = +1, lim inf (a_n) = -1.
- Each subsequence is an increasing sequence converging to L.
- From the proof of monotone convergence theorem, sup(a_n) = lim(a_n) = L. Thus $b_1 = b_2 = \ldots = L$ and lim sup (a_n) = L.
- On the other hand, c_n = a_n. Thus lim inf (a_n) = lim (c_n) = L.
Take the sequence a_n= (0, -1, 0, -2, 0, -3, 0, -4, …). Since the sequence is not lower-bounded, we have lim inf (a_n) = -∞. On the other hand, since each b_n = 0 we have lim sup (a_n) = 0.

Basic Properties.

Pick any sequences (a_n) and (b_n). Then:

$\lim \sup (c\cdot a_n) = c\cdot\lim \sup (a_n)$ if c > 0;
$\lim \sup (c\cdot a_n) = c\cdot\lim \inf (a_n)$ if c < 0;
$\lim \sup (a_n + b_n) \le \lim\sup(a_n) + \lim\sup(b_n)$ . [ If you can’t recall which way the inequality goes, just think of the two alternating sequences (1, 0, 1, 0, 1, 0, …) and (0, 1, 0, 1, 0, 1, …). Then LHS = 1 and RHS = 1+1 = 2. ]

Proof (a bit sketchy).

This follows from $\sup\{c\cdot a_n, c\cdot a_{n+1}, \ldots\} = c\cdot\sup\{a_n, a_{n+1}, \ldots\}$ if c>0.
Follows from $\sup\{c\cdot a_n, c\cdot a_{n+1}, \ldots\} = c\cdot\inf\{a_n, a_{n+1}, \ldots\}$ if c<0.
Let $u_n =\sup\{a_n, a_{n+1}, \ldots\}$ and $v_n=\sup\{b_n, b_{n+1},\ldots \}$ . Then for each m≥n, $a_m\le u_n, b_m\le v_n$ , and thus $a_m+b_m \le u_n+v_n$ . Hence, $\sup\{a_n+b_n, a_{n+1}+b_{n+1},\ldots\} \le u_n+v_n$ . Take the limit of both sides. ♦

Theorem.

Let (a_n) be a sequence.

If lim (a_n) = L (resp. +∞, -∞), then lim inf (a_n) = lim sup (a_n) = L (resp. +∞, -∞).
Conversely, if lim inf (a_n) = lim sup (a_n) = L (resp. +∞, -∞), then lim (a_n) = L (resp. +∞, -∞).

Proof

For the first statement, consider the case of finite limit. Given ε>0, for some N, we have L-(ε/2) < a_n < L+(ε/2) for all n>N. So L+(ε/2) is an upper bound for a_n, a_n+1, a_n+2…. and b_n, being the least upper bound of these values, satisfy b_n ≤ L+(ε/2) < L+ε. Next, since a_n > L-(ε/2), we have b_n ≥ a_n > L-ε. This gives |b_n–L| < ε for all n>N as desired. So lim sup (a_n) = L. The case where lim inf (a_n) = L is similar, or we can just replace (a_n) by (-a_n).

For the second statement, let ε>0. Eventually we have:

$\begin{aligned} L-\epsilon< &\sup\{a_n, a_{n+1}, a_{n+2}, \ldots\} < L+\epsilon,\\L-\epsilon< &\inf\{a_n, a_{n+1}, a_{n+2}, \ldots\}<L+\epsilon.\end{aligned}$

In particular, $a_n \le \sup\{a_n, a_{n+1}, \ldots\} < L+\epsilon$ and $a_n \ge \inf\{a_n, a_{n+1}, \ldots\} > L-\epsilon$ . Hence, eventually $L-\epsilon<a_n < L+\epsilon$ which proves that lim (a_n) = L. ♦