# Recitation 4

Young man, in mathematics you don’t understand things. You just get used to them.
John von Neumann

Problem 1: Suppose $f\colon \mathbb{R} \to \mathbb{R}$ is monotone increasing. Prove that (a) if $f$ is bounded from above, then $\lim_{x\to\infty}f(x)$ exists; (b) if $f$ is not bounded from above, then $\lim_{x\to\infty}f(x)=\infty$.

Idea: (a) Let $L$ be the supremum of the image of $f$. Argue that $\lim_{x\to\infty}f(x) = L$. (b) Write down the statements “$f$ is not bounded from above” and “$\lim_{x\to\infty}f(x)=\infty$” mathematically.

Problem 2: Define mathematically that the sequence $\{a_n\}_{n=1}^\infty$ (a) converges to 5; (b) does not converge to 3; (c) converges to $\infty$; (d) does not converge to $\infty$.

Answer: (a) for every $\epsilon > 0$ there exists $N$ such that $|a_n-5|<\epsilon$ for all $n> N$. (b) there exists $\epsilon > 0$ for every $N$, $|a_n-3|\ge \epsilon$ for some $n>N$. (c) for every $M$, there exists $N$ such that $a_n > M$ for all $n>N$. (d) there exists $M$, for every $N$, $a_n < M$ for some $n > N$.

Problem 3: Prove that if $\{a_n\}_{n=1}^\infty$ converges (in the strict sense), then it is bounded.

Proof: Suppose the sequence converges to $L$. We know from Problem 2(a) there exists $N$ such that $|a_n-L|<1$ for all $n > N$. Therefore the sequence is bounded from below by $\min(a_1, \dots, a_N, L-1)$ and bounded from above by $\max(a_1, \dots, a_N, L+1)$.

Problem 4: Adding, deleting or changing finite number of elements in a sequence does not change its convergence and its limit (in case it converges).

Hint: It is enough to show that adding, deleting or changing one element in a sequence does not change its convergence and its limit.

Fact: Suppose $\{a_n\}_{n=1}^\infty$ converges to $L$, then $\{|a_n|\}_{n=1}^\infty$ converges to $|L|$.

Application of problem 4: Prove that $\{0, 1, 0, 1, \dots \}$ does not converge.

Proof: Suppose $\{a_n\}_{n=1}^\infty = \{0, 1, 0, 1, \dots \}$ converges. By problem 4, $\{a_{n+1}\}_{n=1}^\infty$ converges and has the same limit as $\{a_n\}_{n=1}^\infty$. Therefore $\lim_{n\to\infty}{a_n} = \lim_{n\to\infty}{a_{n+1}}$, and so $\lim_{n\to\infty}|{a_n}-a_{n+1}| = 0$. However, $|a_n - a_{n+1}| = 1$ for all $n$. A contradiction.

Problem 5: Suppose that $\lim_{x\to 0}f(x) = L$. Prove that $\lim_{n\to\infty}f(1/n) = L$.

Proof: Given $\epsilon > 0$. We know that there is $\delta > 0$ such that $|f(x)-L|<\epsilon$ for all $0<|x|<\delta$. Take $N = 1/\delta$. For all $n > N$, we know that $0 < 1/n < 1/N = \delta$, thus $|f(1/n)-L|<\epsilon$.

Application of problem 5: $\lim_{n\to\infty}\sin(1/n) = 0, \lim_{n\to\infty}\cos(1/n) = 0, \lim_{n\to\infty}\sin(1/n^2) = 0$.

Problem 6: Let $\{a_n\}_{n=1}^\infty$ and $\{b_n\}_{n=1}^\infty$ be two sequences. Prove that if both $\{a_n-b_n\}_{n=1}^\infty$ and $\{a_n\}_{n=1}^\infty$ converge, then so does $\{b_n\}_{n=1}^\infty$.

Idea: $b_n = a_n - (a_n - b_n)$.

Problem 7: Find (a) $\lim_{n\to\infty}\frac{n^5-n^4-7n+4}{3n^5-n^3-2n^2-1}$; (b) $\lim_{n\to\infty}\sqrt[n]{1^n + 2^n + \dots + 10^n}$; (c) $\lim_{n\to\infty}\sqrt[n]{1+2+\dots+n}$; (d) $\lim_{n\to\infty}\sqrt[n]{1^5+2^5+\dots+n^5}$; (e) $\lim_{n\to\infty}\sqrt[n]{1(n-1)+2(n-2)+\dots+(n-1)1}$.

Answers: (a) 1/3; (b) 10; (c) 1; (d) 1; (e) 1.