Convergence of $\sum_{n = 1}^\infty 1/n^2$.

Question

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.

Answer 1

This problem is quite notorious specifically because it defies intuition. I agree with you, if we didn't know better, a "good" guess would be that they both converge.

One possible intuitive explanation for why the harmonic series diverges is this:

understand the relation between the harmonic series and the natural log. When we take the sum $1/1+1/2+1/3+\cdots$, we are finding a crude estimate for the integral of $1/x$ from $1$ to infinity. As you probably know, the integral of $1/x$ is the natural log, so we are really taking the limit of $\ln(x)$ as $x$ goes to infinity.

Now it is somewhat easier to see why $\ln(x)$ keeps getting larger. The log of a very large number is asking "$e$ to what power gives this number?" Since we can keep finding larger numbers, there is no bound on $\ln(x)$.

Of course this is intended to be an intuitive connection. You can rigorously prove that the harmonic series diverges, while your other series converges.

Answer 2

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.