Recitation 5

Problem 1: Consider the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_1 = 1$ and $a_{n+1} = a_n + 1/a_n$ for every $n \ge 1$. Show that $\{a_n\}_{n=1}^\infty$ is monotone increasing and $\lim_{n\to\infty} a_n = \infty$.

Proof: One can see that each $a_n$ is positive (a rigorous proof requires mathematical induction). Thus $a_{n+1} = a_n + 1/a_n > a_n$, that is, the sequence is monotone increasing. Suppose, for a moment, that the sequence is bounded from above, by Problem 1(a) from Recitation 4, the sequence has a limit, say $L$. Because $a_na_{n+1} = a_n^2 + 1$, $L^2 = \lim_{n\to\infty}a_na_{n+1} = \lim_{n\to\infty}a_n^2+1 = L^2+1$, which is impossible. Therefore the sequence is not bounded from above. By Problem 1(b) from Recitation 4, $\lim_{n\to\infty} a_n = \infty$.

Remark: Alternatively, one can prove by mathematical induction that $a_n \le n$. Again, by induction, we can show that $a_n \ge 1 + 1/1 + 1/2 + \dots + 1/(n-1)$. Denote by $H_n := 1/1 + 1/2 + \dots + 1/n$, which is the Harmonic series. It can be shown that $H_n \ge 1 + \lfloor \log_2 n \rfloor / 2$, which diverges to infinity.

Problem 2: Suppose that $\lim_{n\to\infty} a_n = 0$. Prove that $\lim_{n\to\infty} 5^{a_n} = 1$.

Proof: We know that $\lim_{n\to\infty}5^{1/n} = 1$. Given $\epsilon > 0$, find $M$ such that $5^{1/M} < 1+\epsilon$. Since $\lim_{n\to\infty} a_n = 0$, we can find $N$ such that $-1/M < a_n < 1/M$ for all $n > N$. Now, for all $n > N$, $1 - \epsilon < \frac{1}{1+\epsilon} < 5^{-1/M} < 5^{a_n} < 5^{1/M} < 1 + \epsilon$, that is, $|5^{a_n} - 1| < \epsilon$.

Problem 3: Prove that if $\lim_{n\to\infty} a_n = A > 0$ and $\lim_{n\to\infty} b_n = B$, then $\lim_{n\to\infty} a_n^{b_n} = A^B$.

Exercise 1: Use mathematical induction to prove the following. For all $\epsilon\in(0,1)$ and $M\in\mathbb{N}$, $(1-\epsilon)^M \ge 1-M\epsilon$.

Exercise 2: Suppose that $\epsilon \in (0, 1)$, $M\in\mathbb{N}$. If $1-\epsilon/M and $-M < b < M$, then $1 - \epsilon < a^b < 1+\epsilon$.

Proof: We first prove that case when $A = 1$. Since $\lim_{n\to\infty} b_n = B$, by Problem 3 from Recitation 4, the sequence $\{b_n\}_{n=1}^\infty$ is bounded. Let $N$ be a natural number such that $-M < b_n < M$ for all $n$. Suppose that $\epsilon \in (0,1)$ is given. Since $\lim_{n\to\infty} a_n = 1$, there exists $N$ such that $1-\epsilon/M. For all $n > N$, by Exercise 2, $1 - \epsilon < a_n^{b_n} < 1 + \epsilon$, and so $\lim_{n\to\infty} a_n^{b_n} = 1$.

Finally, we prove the general case. Let $a'_n = a_n / A$. Because $\lim_{n\to\infty} a_n' = \lim_{n\to\infty} a_n/A = 1$, we know that $\lim_{n\to\infty} a_n'^{b_n} = 1$. Moreover, from problem 2, we know that $\lim_{n\to\infty} A^{b_n} = A^B$. Therefore $\lim_{n\to\infty} a_n^{b_n} = \lim_{n\to\infty} (a_n' \cdot A)^{b_n} = \lim_{n\to\infty} a_n'^{b_n} \lim_{n\to\infty} A^{b_n} = A^B$.

Facts: If $\lim_{n\to\infty}a_n=\infty$, then $\lim_{n\to\infty}(1+1/a_n)^{a_n}=e$. If $\lim_{x\to x_0}f(x)=\infty$, then $\lim_{n\to\infty}(1+1/f(x))^{f(x)}=e$.

Corollary 1: If $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n/a_n = L$, then (a) $\lim_{n\to\infty}(1+1/a_n)^{b_n}=e^L$ and (b) $\lim_{n\to\infty}(1-1/a_n)^{b_n}=e^{-L}$.

Corollary 2: If $\lim_{x\to x_0}f(x)=\infty$ and $\lim_{x\to 0}g(x)/f(x) = L$, then (a) $\lim_{n\to\infty}(1+1/f(x))^{g(x)}=e^{L}$ and (b) $\lim_{n\to\infty}(1-1/f(x))^{g(x)}=e^{-L}$.

Problem 4: Find the limits (a) $\lim_{n\to\infty} \left(\sin(1/n)+\cos(1/n)\right)^{n^2}$; (b) $\lim_{n\to\infty} \left(1+\tan(1/n)\right)^{n}$; (c) $\lim_{n\to\infty} \left(\frac{n^2+n+1}{n^2-n-1}\right)^{n}$; (d) $\lim_{x\to 0} \left(\cos x\right)^{1/x^2}$.

Answers: (a) $\infty$; (b) $e$; (c) $e^2$; (d) $e^{-1/2}$.

Idea: (a) In Corollary (1a), take $a_n = 1/(\sin(1/n)+\cos(1/n)-1)=1/(2\sin(1/2n)(\cos(1/2n)-\sin(1/2n)))$ and $b_n = n^2$. (b) In Corollary (1a), take $a_n = 1/\tan(1/n)$ and $b_n = n$. (c) In Corollary (1a), take $a_n = (n^2-n-1)/(2n+2)$ and $b_n = n$. (d) In Corollary (2b), take $x_0 = 0$, $f(x) = 1/(1-\cos x) = 1/(2\sin^2(x/2))$ and $g(x) = 1/x^2$.

Problem 5: Let $f\colon \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) = f(2x)$ for every $x$. Show that $f$ is constant.

Idea: In Problem 6 from Homework 2, we actually proved the following. Let $f\colon \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) = f(2x)$ for every $x$. If $\lim_{x\to 0}f(x) = L$, then $f(x) = L$ for all $x \neq 0$. In particular, if $f$ is continuous, $L$ must be $f(0)$.