Why does the series $\sum_{n=1}^\infty\frac1n$ not converge?

Let's group the terms as follows:

Group $1$ : $\displaystyle\frac11\qquad$ ($1$ term)

Group $2$ : $\displaystyle\frac12+\frac13\qquad$($2$ terms)

Group $3$ : $\displaystyle\frac14+\frac15+\frac16+\frac17\qquad$($4$ terms)

Group $4$ : $\displaystyle\frac18+\frac19+\cdots+\frac1{15}\qquad$ ($8$ terms)

$\quad\vdots$

In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $\dfrac1{2^n}$. For example all elements in group $2$ are larger than $\dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} \cdot \dfrac1{2^n} = \dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $\dfrac1{2}$, it follows that the total sum is infinite.

This proof is often attributed to Nicole Oresme.

There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.

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EDIT

It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.

This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $$e^{H_n}=\prod_{k=1}^n e^{1/k}\gt\prod_{k=1}^n\left(1+\frac{1}{k}\right)=n+1.$$ Therefore, $H_n\gt\log(n+1)$, so we are done. We used $e^x\gt1+x$ and telescoped the resulting product.