§12 lim sup's and lim inf's

Let $(s_n)$ be any sequence of real numbers and $S$ its set of subsequential limits. Recall

$$\limsup s_n = \lim_{N\to\infty} \sup\{s_n : n > N\} = \sup S \tag{$*$}$$ $$\liminf s_n = \lim_{N\to\infty} \inf\{s_n : n > N\} = \inf S. \tag{$**$}$$

The first equalities are Definition 10.6 and the second are Theorem 11.8. This section is meant to build familiarity with these concepts; most of the material is in the exercises. We illustrate the techniques by proving some results needed later.

Intuitionthe two-sided test for lim sup

When $L = \limsup s_n$ is finite, it is pinned down by two statements, proved in Example 2(a) below:

• for every $\alpha > L$, the terms are eventually below $\alpha$: only finitely many $s_n$ exceed $\alpha$;
• for every $\beta < L$, the terms are above $\beta$ infinitely often.

So $\limsup s_n$ is the dividing line between levels that the sequence eventually stays below and levels that it keeps rising above. Most proofs about $\limsup$ come down to using one of these two statements.

Terms shown60

sₙv_N = sup{sₙ : n > N}u_N = inf{sₙ : n > N}

The values 1/2, −1/2, −1, −1/2, 1/2, 1 repeat forever. Subsequential limits: {−1, −1/2, 1/2, 1}. lim sup = 1, lim inf = −1.

At N = 60: v_N ≈ 1.0000 and u_N ≈ -1.0000. As N grows, v_N decreases toward lim sup sₙ and u_N increases toward lim inf sₙ.

Another view: pick terms yourself and see which subsequential limit they approach. $\limsup s_n$ and $\liminf s_n$ are the extreme ones.

Even nOdd nEvery 3rdn ≡ 1 mod 6Clear

Choose the terms of your subsequence (n₁ < n₂ < …)

123456789101112131415161718192021222324252627282930313233343536

selected terms s_(nₖ)lim sup = 1lim inf = -1other subsequential limits

✗ The later selected terms are spread over a range of 1.5, so this selection does not settle down. It is not a convergent subsequence (within this window).

The six values repeat, so each value is a subsequential limit (pick every sixth term from the right starting point). The set is {−1, −1/2, 1/2, 1}.

Tab to a term’s checkbox and press Space to add or remove it, or click a point on the plot. Verdicts look only at the first 36 terms.

12.1 Theorem.

If $(s_n)$ converges to a positive real number $s$ and $(t_n)$ is any sequence, then

$$\limsup s_n t_n = s \cdot \limsup t_n.$$

Here we allow the conventions $s \cdot (+\infty) = +\infty$ and $s \cdot (-\infty) = -\infty$ for $s > 0$.

Proof

First we show

$$\limsup s_n t_n \ge s \cdot \limsup t_n. \tag{1}$$

Let $\beta = \limsup t_n$.

Case 1: $\beta$ finite. By Theorem 11.7 there is a subsequence with $\lim_k t_{n_k} = \beta$. Also $\lim_k s_{n_k} = s$ (Theorem 11.3), so $\lim_k s_{n_k} t_{n_k} = s\beta$. This is a subsequential limit of $(s_n t_n)$, and $\limsup s_n t_n$ is the largest one, so $s\beta \le \limsup s_n t_n$.

Case 2: $\beta = +\infty$. Some subsequence has $\lim_k t_{n_k} = +\infty$. Since $\lim_k s_{n_k} = s > 0$, Theorem 9.9 gives $\lim_k s_{n_k} t_{n_k} = +\infty$, so $\limsup s_n t_n = +\infty$.

Case 3: $\beta = -\infty$. The right side of (1) is $-\infty$, so (1) is obvious.

For the reverse inequality, a little trick. Ignoring finitely many terms we may assume all $s_n \ne 0$, so $\lim \frac1{s_n} = \frac1s$ by Lemma 9.5. Apply (1) with $\frac1{s_n}$ in place of $s_n$ and $s_n t_n$ in place of $t_n$:

$$\limsup t_n = \limsup\left(\frac1{s_n}\right)(s_n t_n) \ge \frac1s \limsup s_n t_n,$$

i.e. $\limsup s_n t_n \le s \cdot \limsup t_n$.

∎

Intuitionthe trick in the second half

The reverse inequality is obtained by applying the first half to a cleverly chosen pair of sequences. Proving one inequality and then getting the other from it by substitution, here $\frac{1}{s_n}$ for $s_n$, is a pattern worth recognizing.

Example 1.

The hypothesis $s > 0$ cannot be relaxed to $s = 0$. Take $s_n = -\frac1n$ and $t_n = -n^2$. Then $s_n t_n = n$, so $\limsup s_n t_n = +\infty$, while $s \cdot \limsup t_n$ would be $0 \cdot (-\infty)$. We don’t define that, and we certainly wouldn’t define it to be $+\infty$.

The next theorem is used for infinite series, in the proof of the Ratio Test 14.8.

12.2 Theorem.

Let $(s_n)$ be any sequence of nonzero real numbers. Then

$$\liminf \left|\frac{s_{n+1}}{s_n}\right| \le \liminf |s_n|^{1/n} \le \limsup |s_n|^{1/n} \le \limsup \left|\frac{s_{n+1}}{s_n}\right|.$$

Used in Corollary 12.3, Exercise 12.11, Theorem 14.9

Proof

The middle inequality is obvious, and the outer ones have similar proofs. We prove the third; the first is Exercise 12.11.

Let $\alpha = \limsup |s_n|^{1/n}$ and $L = \limsup\left|\frac{s_{n+1}}{s_n}\right|$. We must show $\alpha \le L$. This is obvious if $L = +\infty$, so assume $L < +\infty$. It suffices to show

$$\alpha \le L_1 \quad\text{for any } L_1 > L. \tag{1}$$

Since $L = \lim_{N\to\infty} \sup\left\{\left|\frac{s_{n+1}}{s_n}\right| : n > N\right\} < L_1$, there is $N$ with $\sup\left\{\left|\frac{s_{n+1}}{s_n}\right| : n \ge N\right\} < L_1$, so

$$\left|\frac{s_{n+1}}{s_n}\right| < L_1 \quad\text{for } n \ge N. \tag{2}$$

For $n > N$ write $|s_n| = \left|\frac{s_n}{s_{n-1}}\right| \cdot \left|\frac{s_{n-1}}{s_{n-2}}\right| \cdots \left|\frac{s_{N+1}}{s_N}\right| \cdot |s_N|$. There are $n - N$ fractions, so by (2), $|s_n| < L_1^{n-N}|s_N|$. With the constant $a = L_1^{-N}|s_N| > 0$ this says $|s_n| < L_1^n a$, so $|s_n|^{1/n} < L_1 a^{1/n}$ for $n > N$. Since $\lim a^{1/n} = 1$ (Theorem 9.7(d)), we conclude $\alpha = \limsup|s_n|^{1/n} \le L_1$ (see Exercise 12.1).

∎

Intuition

If the ratios $\left|\frac{s_{n+1}}{s_n}\right|$ are eventually below $L_1$, then each step multiplies the size by less than $L_1$, so $|s_n|$ shrinks at least as fast as a constant times $L_1^n$. Taking $n$th roots strips off the constant, because $a^{1/n} \to 1$. Since $|s_n| = |s_1| \cdot \left|\frac{s_2}{s_1}\right| \cdots \left|\frac{s_n}{s_{n-1}}\right|$, the root $|s_n|^{1/n}$ is essentially the geometric average of the growth factors, while the ratio measures each individual step. An average can never be more extreme than the individual steps, which is what Theorem 12.2 says, and why the Root Test sees at least as much as the Ratio Test in §14.

12.3 Corollary.

If $\lim\left|\frac{s_{n+1}}{s_n}\right|$ exists and equals $L$, then $\lim |s_n|^{1/n}$ exists and equals $L$.

Used in Exercise 12.14, Remark 14.10

Proof

All four quantities in Theorem 12.2 equal $L$. Apply Theorem 10.7.

∎

For example, with $s_n = n$ the ratios $\frac{n+1}{n} \to 1$, which gives another proof that $\lim n^{1/n} = 1$.

Example 2.

1. If $L = \limsup s_n \ne +\infty$, then for every $\alpha > L$ the set $\{n : s_n > \alpha\}$ is finite. Otherwise $\sup\{s_n : n > N\} > \alpha$ for all $N$, so $L \ge \alpha > L$. If $L \ne -\infty$, then for every $\beta < L$ the set $\{n : s_n > \beta\}$ is infinite. Otherwise $s_n \le \beta$ for $n \ge N_0$, so $\limsup s_n \le \beta < L$.
2. The set $\{n : s_n > \limsup s_n\}$ can be infinite: take $s_n = \frac1n$. So can $\{n : s_n < \liminf s_n\}$: take $s_n = -\frac1n$.
3. If $L_0 = \liminf t_n \ne -\infty$, then $\{n : t_n < \beta_0\}$ is finite for $\beta_0 < L_0$. If $L_0 \ne +\infty$, then $\{n : t_n < \alpha_0\}$ is infinite for $\alpha_0 > L_0$. This follows from (a) applied to $s_n = -t_n$, since $\liminf t_n = -\limsup(-t_n)$ (Exercise 11.8).
4. If $\liminf s_n < \limsup s_n$, the set $\{n : \liminf s_n \le s_n \le \limsup s_n\}$ can be empty. For example, $s_n = (-1)^n\left(1 + \frac1n\right)$.

Common pitfall

• $\limsup s_n$ is usually not $\sup\{s_n\}$. The supremum can be set by a single early term, while $\limsup$ ignores any finite number of terms.
• Just as for suprema (§4), strict inequalities can be lost: if $s_n < \alpha$ for every $n$, you only get $\limsup s_n \le \alpha$. Take $s_n = 1 - \frac1n$ and $\alpha = 1$.
• $\limsup$ does not split over sums. The quiz below gives a sequence where $\limsup(s_n + t_n) < \limsup s_n + \limsup t_n$.

Reading alongside Abbott

Abbott introduces the limit superior only in Exercise 2.4.7, as the limit of the suprema of the tails (Definition 10.6 here). Abbott’s Ratio Test (Exercise 2.7.9) is stated with an ordinary limit of ratios, so it never needs Theorem 12.2. The $\limsup$ versions in §14 are stronger, because they apply even when the ratios have no limit.

Check your understanding

Take the quiz (3 questions)

Let $s_n = \frac{n^2}{2^n}$. What is $\lim s_n^{1/n}$?

$\frac12$$1$$2$

Let $s_n = (-1)^n$ and $t_n = (-1)^{n+1}$. How does $\limsup(s_n + t_n)$ compare with $\limsup s_n + \limsup t_n$?

$0 < 2$: the inequality can be strict.They are equal; $\limsup$ is additive.$2 > 0$: the first is larger.

If $\lim s_n = -1$, what is $\limsup s_n t_n$?

$-\liminf t_n$$-\limsup t_n$$\limsup t_n$

Exercises

Exercise 12.1

Let $(s_n)$ and $(t_n)$ be sequences with $s_n \le t_n$ for all $n > N_0$. Show $\liminf s_n \le \liminf t_n$ and $\limsup s_n \le \limsup t_n$.

Hint

Use Definition 10.6 and Exercise 9.9(c).

Solution

For every $N \ge N_0$ and every $n > N$ we have $s_n \le t_n$. So any upper bound of $\{t_n : n > N\}$ is an upper bound of $\{s_n : n > N\}$, which gives $\sup\{s_n : n > N\} \le \sup\{t_n : n > N\}$, and in the same way $\inf\{s_n : n > N\} \le \inf\{t_n : n > N\}$. Both sides of each inequality have limits as $N \to \infty$ (the sups decrease and the infs increase), so Exercise 9.9(c) gives $\limsup s_n \le \limsup t_n$ and $\liminf s_n \le \liminf t_n$.

Tags: Standardlim sup & lim inf

Used in Theorem 12.2, Exercise 12.11

Exercise 12.2

Prove $\limsup |s_n| = 0$ if and only if $\lim s_n = 0$.

Hint

$0 \le \liminf|s_n| \le \limsup |s_n|$. Use Theorem 10.7 and Exercise 8.6(a).

Solution

($\Rightarrow$) Since $|s_n| \ge 0$, every infimum of a tail is $\ge 0$, so $0 \le \liminf|s_n| \le \limsup|s_n| = 0$. Thus $\liminf|s_n| = \limsup|s_n| = 0$, and by Theorem 10.7 $\lim|s_n| = 0$, which gives $\lim s_n = 0$ by Exercise 8.6(a).

($\Leftarrow$) If $\lim s_n = 0$ then $\lim|s_n| = 0$ by Exercise 8.6(a), and for a convergent sequence $\limsup$ equals the limit (Theorem 10.7). So $\limsup|s_n| = 0$.

Tags: Standardlim sup & lim inf

Exercise 12.3

Let $(s_n)$ and $(t_n)$ repeat in cycles of four:

$$(s_n) = (0, 1, 2, 1,\ 0, 1, 2, 1,\ \dots), \qquad (t_n) = (2, 1, 1, 0,\ 2, 1, 1, 0,\ \dots).$$

Find

1. $\liminf s_n + \liminf t_n$
2. $\liminf(s_n + t_n)$
3. $\liminf s_n + \limsup t_n$
4. $\limsup(s_n + t_n)$
5. $\limsup s_n + \limsup t_n$
6. $\liminf(s_n t_n)$
7. $\limsup(s_n t_n)$

Solution

$s_n + t_n$ cycles through $2, 2, 3, 1$ and $s_n t_n$ through $0, 1, 2, 0$.

1. $0$
2. $1$
3. $2$
4. $3$
5. $4$
6. $0$
7. $2$

Tags: Warm-uplim sup & lim infComputation

Exercise 12.4

Show $\limsup(s_n + t_n) \le \limsup s_n + \limsup t_n$ for bounded sequences $(s_n)$ and $(t_n)$.

Hint

First show $\sup\{s_n + t_n : n > N\} \le \sup\{s_n : n > N\} + \sup\{t_n : n > N\}$, then apply Exercise 9.9(c).

Solution

Fix $N$. For $n > N$, $s_n + t_n \le \sup\{s_k : k > N\} + \sup\{t_k : k > N\}$, so the left side’s supremum satisfies $\sup\{s_n + t_n : n > N\} \le \sup\{s_n : n > N\} + \sup\{t_n : n > N\}.$ As $N \to \infty$ the right side tends to $\limsup s_n + \limsup t_n$ (Theorem 9.3), and the left side to $\limsup(s_n + t_n)$. Exercise 9.9(c) gives the inequality.

Tags: Standardlim sup & lim inf

Used in Exercise 12.5

Exercise 12.5

Use Exercises 11.8 and 12.4 to prove $\liminf(s_n + t_n) \ge \liminf s_n + \liminf t_n$ for bounded sequences $(s_n)$ and $(t_n)$.

Solution

By Exercise 11.8 and then 12.4 applied to $(-s_n)$ and $(-t_n)$, $\liminf(s_n + t_n) = -\limsup(-s_n - t_n) \ge -\bigl[\limsup(-s_n) + \limsup(-t_n)\bigr] = \liminf s_n + \liminf t_n .$

Tags: Standardlim sup & lim inf

Exercise 12.6

Let $(s_n)$ be a bounded sequence and $k \ge 0$.

1. Prove $\limsup(ks_n) = k \cdot \limsup s_n$.
2. Do the same for $\liminf$. (Use Exercise 11.8.)
3. What happens in (a) and (b) if $k < 0$?

Solution

(c) They swap: $\limsup(ks_n) = k \cdot \liminf s_n$ and $\liminf(ks_n) = k \cdot \limsup s_n$.

Tags: Standardlim sup & lim inf

Exercise 12.7

Prove that if $\limsup s_n = +\infty$ and $k > 0$, then $\limsup(ks_n) = +\infty$.

Solution

The tail suprema $\sup\{s_n : n > N\}$ decrease in $N$ and tend to $+\infty$, so each of them is $+\infty$. For $k > 0$, $\sup\{ks_n : n > N\} = k\sup\{s_n : n > N\} = +\infty$ for every $N$, so $\limsup(ks_n) = +\infty$.

Tags: Warm-uplim sup & lim inf

Used in Exercise 12.9

Exercise 12.8

Let $(s_n)$ and $(t_n)$ be bounded sequences of nonnegative numbers. Prove $\limsup s_n t_n \le (\limsup s_n)(\limsup t_n)$.

Hint

$\sup\{s_n t_n : n > N\} \le \sup\{s_n : n > N\} \cdot \sup\{t_n : n > N\}$ because all terms are nonnegative.

Solution

Let $A_N = \sup\{s_n : n > N\}$ and $B_N = \sup\{t_n : n > N\}$; both are $\ge 0$. For $n > N$, $0 \le s_n \le A_N$ and $0 \le t_n \le B_N$, so $s_nt_n \le A_NB_N$, and hence $\sup\{s_nt_n : n > N\} \le A_NB_N$. As $N \to \infty$, $A_N \to \limsup s_n$ and $B_N \to \limsup t_n$, so $A_NB_N \to (\limsup s_n)(\limsup t_n)$ by Theorem 9.4. Exercise 9.9(c) gives the claim.

Tags: Challengelim sup & lim inf

Exercise 12.9

1. Prove that if $\lim s_n = +\infty$ and $\liminf t_n > 0$, then $\lim s_n t_n = +\infty$.
2. Prove that if $\limsup s_n = +\infty$ and $\liminf t_n > 0$, then $\limsup s_n t_n = +\infty$.
3. Observe that Exercise 12.7 is the special case of (b) where $t_n = k$ for all $n$.

Solution

In (a) and (b), pick a real $m$ with $0 < m < \liminf t_n$. Since $\inf\{t_n : n > N\} \to \liminf t_n$, there is $N_0$ with $\inf\{t_n : n > N_0\} > m$, i.e. $t_n > m$ for all $n > N_0$.

1. Let $M > 0$. Since $s_n \to +\infty$, there is $N_1 \ge N_0$ with $s_n > M/m > 0$ for $n > N_1$, and then $s_nt_n > m s_n > M$.
2. Let $M > 0$ and $N \ge N_0$. Because $\limsup s_n = +\infty$, every tail of $(s_n)$ is unbounded above, so some $n > N$ has $s_n > M/m > 0$; for that $n$, $s_nt_n > ms_n > M$. Thus $\sup\{s_nt_n : n > N\} > M$ for all $N \ge N_0$ and all $M$, so these suprema are $+\infty$ and $\limsup s_nt_n = +\infty$.
3. With $t_n = k > 0$ we have $\liminf t_n = k > 0$, so (b) gives $\limsup(ks_n) = +\infty$.

Tags: Standardlim sup & lim inf

Exercise 12.10

Prove $(s_n)$ is bounded if and only if $\limsup |s_n| < +\infty$.

Solution

($\Rightarrow$) If $|s_n| \le M$ for all $n$, then every tail supremum is at most $M$, so $\limsup|s_n| \le M < +\infty$.

($\Leftarrow$) Suppose $L = \limsup|s_n| < +\infty$; note $L \ge 0$. By Example 2(a), with $\alpha = L + 1$, only finitely many $n$ satisfy $|s_n| > L + 1$. So $|s_n| \le \max\{L + 1, \text{those finitely many } |s_n|\}$ for all $n$, and $(s_n)$ is bounded.

Tags: Standardlim sup & lim inf

Exercise 12.11

Prove the first inequality in Theorem 12.2.

Hint

Mirror the proof of the third inequality: take $L_1 < \liminf\left|\frac{s_{n+1}}{s_n}\right|$ and get $|s_n| > L_1^{n-N}|s_N|$.

Solution

Let $\ell = \liminf\left|\frac{s_{n+1}}{s_n}\right|$ and $\beta = \liminf|s_n|^{1/n}$; we show $\ell \le \beta$. Since both are $\ge 0$ this is clear if $\ell = 0$, so let $\ell > 0$. It suffices to show $\beta \ge \ell_1$ for every real $\ell_1$ with $0 < \ell_1 < \ell$.

By Example 2(c), $\left|\frac{s_{n+1}}{s_n}\right| > \ell_1$ for all $n \ge N$ for some $N$. By induction, $|s_n| > \ell_1^{\,n-N}|s_N|$ for $n > N$, so $|s_n|^{1/n} > \ell_1 \cdot c^{1/n}, \qquad c = \ell_1^{-N}|s_N| > 0 .$ Since $c^{1/n} \to 1$ (Theorem 9.7(d)), the right side tends to $\ell_1$. By Exercise 12.1, $\beta = \liminf|s_n|^{1/n} \ge \lim \ell_1c^{1/n} = \ell_1$.

Tags: Standardlim sup & lim inf

Used in Theorem 12.2

Exercise 12.12

Let $(s_n)$ be a sequence of nonnegative numbers, and let $\sigma_n = \frac1n(s_1 + s_2 + \cdots + s_n)$.

1. Show $\liminf s_n \le \liminf \sigma_n \le \limsup \sigma_n \le \limsup s_n$.
2. Show that if $\lim s_n$ exists, then $\lim \sigma_n$ exists and $\lim \sigma_n = \lim s_n$.
3. Give an example where $\lim \sigma_n$ exists, but $\lim s_n$ does not.

Hint

For the last inequality in (a), show first that $M > N$ implies $\sup\{\sigma_n : n > M\} \le \frac1M(s_1 + s_2 + \cdots + s_N) + \sup\{s_n : n > N\}$. For (c), try $s_n = 1 + (-1)^n$.

Solution

1. The middle inequality is automatic. For the last one, let $L = \limsup s_n$; if $L = +\infty$ there is nothing to prove, so let $L < \infty$ and take any $\alpha > L$ (note $\alpha > 0$ since $s_n \ge 0$ forces $L \ge 0$). Choose $N$ with $s_n < \alpha$ for $n > N$ (Example 2(a)) and put $C = s_1 + \cdots + s_N \ge 0$. For $n > M > N$, $\sigma_n = \frac{C}{n} + \frac{s_{N+1} + \cdots + s_n}{n} \le \frac CM + \frac{(n-N)\alpha}{n} \le \frac CM + \alpha,$ so $\sup\{\sigma_n : n > M\} \le \frac CM + \alpha$. Letting $M \to \infty$ gives $\limsup\sigma_n \le \alpha$, and since $\alpha > L$ was arbitrary, $\limsup\sigma_n \le L$. For the first inequality, let $\ell = \liminf s_n \ge 0$. If $\ell = 0$ it is trivial because $\sigma_n \ge 0$. Otherwise take any real $\beta$ with $0 < \beta < \ell$ and $N$ with $s_n > \beta$ for $n > N$ (Example 2(c)). For $M > N$ and $n > M$, $\sigma_n \ge \frac{(n-N)\beta}{n} = \beta - \frac{N\beta}{n} \ge \beta - \frac{N\beta}{M}$, so $\inf\{\sigma_n : n > M\} \ge \beta - \frac{N\beta}M$. Letting $M \to \infty$ gives $\liminf\sigma_n \ge \beta$, hence $\liminf\sigma_n \ge \ell$ (also when $\ell = +\infty$).
2. If $\lim s_n$ exists then $\liminf s_n = \limsup s_n = \lim s_n$ (Theorem 10.7), and (a) squeezes $\liminf\sigma_n$ and $\limsup\sigma_n$ to the same value. So $\lim\sigma_n$ exists and equals $\lim s_n$.
3. Let $s_n = 1 + (-1)^n$, which alternates $0, 2, 0, 2, \dots$ and has no limit. Among $s_1, \dots, s_n$ there are $\lfloor n/2\rfloor$ twos, so $\sigma_n = \frac{2\lfloor n/2\rfloor}{n} \to 1$.

Tags: Challengelim sup & lim inf

Exercise 12.13

Let $(s_n)$ be a bounded sequence in $\mathbb{R}$. Let $A$ be the set of $a \in \mathbb{R}$ such that $\{n : s_n < a\}$ is finite, i.e. all but finitely many $s_n$ are $\ge a$. Let $B$ be the set of $b \in \mathbb{R}$ such that $\{n : s_n > b\}$ is finite. Prove $\sup A = \liminf s_n$ and $\inf B = \limsup s_n$.

Hint

Use Example 2(a) and (c).

Solution

Let $\ell = \liminf s_n$ and $L = \limsup s_n$, both finite since $(s_n)$ is bounded. Note that $A$ is closed downward ($a' < a \in A$ implies $\{s_n < a'\} \subseteq \{s_n < a\}$ is finite), and $B$ is closed upward.

$\sup A = \ell$. If $\beta < \ell$, then $\{n : s_n < \beta\}$ is finite (Example 2(c)), so $\beta \in A$; hence $\sup A \ge \ell$. If $\alpha > \ell$, then $\{n : s_n < \alpha\}$ is infinite (Example 2(c)), so $\alpha \notin A$; hence no element of $A$ exceeds $\ell$, and $\sup A \le \ell$.

$\inf B = L$. If $\alpha > L$ then $\{n : s_n > \alpha\}$ is finite (Example 2(a)), so $\alpha \in B$ and $\inf B \le L$. If $\beta < L$ then $\{n : s_n > \beta\}$ is infinite (Example 2(a)), so $\beta \notin B$; hence every element of $B$ is $\ge L$ and $\inf B \ge L$.

Tags: Challengelim sup & lim infSup & inf

Exercise 12.14

Calculate

1. $\lim (n!)^{1/n}$
2. $\lim \frac1n (n!)^{1/n}$

Hint

Apply Corollary 12.3. For (b), with $s_n = \frac{n!}{n^n}$ the ratio is $\frac{s_{n+1}}{s_n} = \left(\frac{n}{n+1}\right)^n = \frac{1}{(1 + 1/n)^n}$.

Solution

(a) $+\infty$, since $\frac{(n+1)!}{n!} = n + 1 \to +\infty$. (b) $\frac1e$.

Tags: Challengelim sup & lim infComputation