Prove formally that a sequence $a_n = 1/n$ converges to $0$. Prove formally that a sequence $a_n = 1/n$ converges to $0$ using formal definition of continuity. 
 A: The sequence $\{a_n\}_{n=1}^\infty$, where $a_n$ is defined as $a_n=\frac1n$, is a convergent sequence. You can prove the fact that it is convergent with a limit of $L$ by showing that

For every $\epsilon > 0$, there exists some $N\in\mathbb N$ such that for every $n>N$, the inequality $|a_n-L|<\epsilon$


On the other hand, the series
$$\sum_{i=1}^\infty \frac 1n$$ is divergent. This is because the sequence $\{S_n\}_{n=1}^\infty$, defined as $$S_n = \sum_{i=1}^n \frac 1i$$ is divergent.
A: To say that $1/n$ converges to $0$ is to say that you can make the difference
$$\vert \frac{1}{n} -0 \vert = \vert \frac{1}{n} \vert = \frac{1}{n}$$
less than any $\epsilon$, no matter how small, just by taking a big enough $n$, going far enough in the sequence.
Now if $\epsilon$ is arbitrary, what $n$ do you have to choose to make sure that
$$\frac{1}{n} < \epsilon \quad \text{?}$$
A: Proof
For any $\varepsilon>0$, there exists a $N=[1/\varepsilon]+1$ (where $[\cdots]$ denotes the floor integer function) such that $$\big|\frac{1}{n}-0\big|=\frac{1}{n}<\frac{1}{N}=\frac{1}{[1/\varepsilon]+1}<\frac{1}{1/\varepsilon}=\varepsilon$$ when $n>N.$ Thus, by the definition of the sequence limit, we are done.
