108學年度｜微積分模組班｜Class 01｜Homework 1.

Homework 1

• Date: 108/09/17 (Week 02)
• Place: 新生教學樓 301

Problem 1.

• Prove that $\displaystyle \lim_{n \to \infty} \dfrac{1}{n^2} = 0$ by using the definition of limits.

Solution.

• Given $\epsilon >0$, take $n_0 = \left\lceil \dfrac{1}{\sqrt{\epsilon}} \right\rceil + 1$, then: $$\left|\dfrac{1}{n^2} - 0\right| = \dfrac{1}{n^2} \leqslant \dfrac{1}{\left(\left\lceil \dfrac{1}{\sqrt{\epsilon}} \right\rceil + 1\right)^2} < \epsilon \text{ whenever } n \geqslant n_0$$ Hence by definition, $\displaystyle \lim_{n \to \infty} \dfrac{1}{n^2} = 0$.

Problem 2.

• Consider the sequence $\{a_n\}_{n = 1}^\infty$ such that $a_n = (−1)^n$. Show that the sequence has no limit.

Solution.

• Suppose the limit exist, say, $L$. (first notice that $L \neq 1$ and $L \neq -1$ in advance) That is, for every $\epsilon > 0$, there exist $n_0 \in \mathbb{N}$ such that $|a_n - L|<\epsilon$ when $n \geqslant n_0$. In particular, take $\epsilon = \min \{|L-1|,|L-(-1)|\} > 0$, then we have $|a_n - L| \geqslant \epsilon$ for all $n \in \mathbb{N}$ (Verify yourself!). Hence, such $n_0 \in \mathbb{N}$ does not exist and thus $\{a_n\}_{n = 1}^\infty$ does not converge.

Problem 3. 

• Prove that the infinite sequence $\{2^n\}_{n \geqslant 1}$ is unbounded.

Solution.

• Suppose it is bounded, that is, there exist $M > 0$ such that $|2^n|<M $ for all $n \in \mathbb{N}$. Obviously, $M > 2$. But note that: $$2^M = (1+1)^M = \sum_{k = 0}^M \binom{M}{k} = 1 + M + \sum_{k = 2}^M \binom{M}{k}>M$$, which leads to a contradiction, hence the set $\{2^n\}_{n \geqslant 1}$ is unbounded.