## 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 (an) be a sequence of real numbers. We write (an) → ∞ if

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

By the same token, (an) → -∞ if

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

[ A sequence which approaches +∞. ]

Let’s define a shorthand here: we say that “X eventually happens” for a sequence (an) if there exists N such that for all nN, X happens for an. 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, Xm eventually happens”, one should not write “Xm 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 (an) → ∞ which says: for every m, eventually an>m. This is different from saying: we eventually have (an>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 (an) → $\left\{\begin{matrix} L\\+\infty\\-\infty\end{matrix}\right\}$ 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

Properties.

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

• If (an) → ∞ or L, and (bn) → ∞, then (an+bn) → ∞.
• If (an) → ∞ or L>0, and (bn) → ∞, then (anbn) → ∞.
• If (an) → -∞ or L<0, and (bn) → ∞, then (anbn) → -∞.
• If (an) → ∞ or -∞, then (1/an) → 0.
• If (an) → 0 for a sequence of positive numbers (an), then (1/an) → ∞.

Proof (Sketch).

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

• we eventually have |anL| < 1, so an>L-1;
• for any M, we eventually have bn>M-(L-1).

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

For the second property: suppose (an) → L>0. Eventually, |anL| < L/2, so an>L/2, and for any M>0, eventually bn>2M/L. Hence, for any M, eventually anbnM. This proves that (anbn) → ∞.

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

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

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

Corollaries

• If (an) → ∞ (resp. -∞), then (-an) → -∞ (resp. ∞). [ Multiply an by the constant sequence of (-1). ]
• If (an) → ∞ or L, and (bn) → -∞, then (anbn) → ∞. [ Write anban+(-bn) and apply two properties. ]

More Properties.

Correspondingly, we now have:

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

Proof.

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

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

## Limits Inferior and Superior

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

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

In summary, we have the following cases:

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

Warning: even if (an) 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 (an) we get three cases.

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

Examples

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

Basic Properties.

Pick any sequences (an) and (bn). 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 mn, $a_m\le u_n, b_m\le v_n$, 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 (an) be a sequence.

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

Proof

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

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\}

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 which proves that lim (an) = L. ♦

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