I have been studying Cauchy criterion for sequences, and have come across a rather simple proof for the harmonic series, and why it diverges.   More so, we have the following:  $$\sum_{n=1}^{\infty}\frac 1n=\infty\Rightarrow divergent$$  Here is my simple proof:  Consider the sequence $\left\{a_n\right\}_{n=1}^{\infty}$ such that $a_n=\frac 1n$, then $\forall \epsilon&gt;0,\exists \ N \in\mathbb{R}$ for $m,n\in \mathbb{N}$, such that,  $$m,n&gt;N\Rightarrow|a_m-a_n|&lt;\epsilon$$  Pick $m=2n$, then we have the following:  $$|a_{2n}-a_n|=\sum_{k=n+1}^{2n}\frac {1}{k}\geq \sum_{k=n+1}^{2n} \frac {1}{2n}=\frac 12$$  So pick $\epsilon=\frac 12\Rightarrow  \left\{a_n\right\}_{n=1}^{\infty}$ is not Cauchy and thus divergent.  My question is, are there any other slick and easy proofs for the above claim, and if so, what are they? This series at first surprised me, as it initially doesn't seem divergent.  Thanks in advance!
Can someone give a simple explanation as to why the       $$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$   doesn't converge, on the other hand it grows very slowly?   I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.
An infinite subset of a countable set is countable In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid?  Let $S$ be an infinite subset of a countable set $T$. Since $T$ is countable, there exists a bijection $f: \mathbb{N} \rightarrow T$. And $S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}$. Let $n_{1}$ be the smallest positive integer such that $f(n_{1}) \in S$. And continue where $n_{k}$ is the smallest positive integer greater than $n_{k-1}$ such that $f(n_{k}) \in S$. And because $S$ is infinite, we continue forever. Now consider the function $\beta : \mathbb{N} \rightarrow S$ which sends $k \rightarrow f(n_{k})$.   So in order for $S$ to be countable, $\beta$ would have to be a bijection. $\beta$ is injective since if $f(n_{k}) = f(n_{j})$, then $n_{k} = n_{j}$ because $f$ is injective. 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