Prove formally that a sequence $a_n = 1/n$ converges to $0$ using formal definition of continuity.

To say that $1/n$ converges to $0$ is to say that you can make the difference

$$vert frac1n -0 vert = vert frac1n vert = frac1n$$

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
$$frac1n < epsilon quad text?$$

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 $Ninmathbb N$ 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.

Proof

For any $varepsilon>0$, there exists a $N=[1/varepsilon]+1$ (where $[cdots]$ denotes the floor integer function) such that $$big|frac1n-0big|=frac1n<frac1N=frac1[1/varepsilon]+1<frac11/varepsilon=varepsilon$$ when $n>N.$ Thus, by the definition of the sequence limit, we are done.