Convergence of $\sum_{n = 1}^\infty 1/n^2$. I know that $\sum_{n=1}^\infty 1/n$ diverges whereas $\sum_{n=1}^\infty 1/n^2$ converges.
Intuitively, I do not see the difference. If $n \to \infty$, the denominators in both fractions will be so big that the fraction approaches zero. So why doesn't both the series converge against zero?
I have read something about it, but I do not see the logic in it, and I guess it's quite obvious, so I hope someone can bring a good explanation.

To see that $\displaystyle\sum_{n=1}^\infty \frac{1}{n}$ diverges, you can group the terms,
$$\begin{alignat*}{2}
\frac{1}{2} + \frac{1}{3} &> \frac{1}{2}+\frac{1}{2} &= 1\\
\frac{1}{4} + \dots +\frac{1}{7} &> \frac{1}{4} + \dots + \frac{1}{4} &= 1\\
&\;\;\vdots&\vdots\;\;\;\,\\
\frac{1}{2^n}+\dots+\frac{1}{2^{n+1}-1} &> \frac{1}{2^n}+\dots +\frac{1}{2^n} &= 1
\end{alignat*}$$
And $\displaystyle\sum_{n=2}^\infty \frac{1}{n^2}$ converges by comparison:
$$
\sum_{n=2}^N\frac{1}{n^2} < \sum_{n=2}^N \frac{1}{n(n-1)} = \sum_{n=2}^N\frac{1}{n-1}-\frac{1}{n} = 1 -\frac{1}{N}\to 1 \quad(\text{telescoping}).
$$
These are the most intuitive explanations I've got.
