I understand the rules of convergence of a series so that I know that $\sum \frac 1 n$ (the harmonic series) diverges and $\sum \frac 1 {n^2}$ squared converges. It doesn't make sense graphically to me. One just gets to $0$ a little faster. What graphically sets that borderline?

For that matter, what makes the harmonic series diverge by the integral test and the alternating harmonic series converge by the alternating series test. That also seems counterintuitive graphically.


The Cauchy condensation test may help here. For a non-increasing sequence $(f(n))_{n\in\mathbb{N}}$ of non-negative real numbers we have (from Oresme's proof of the divergence of the harmonic series) : \begin{align} \tag{1}\sum_{k=1}^\infty f(k)&\le \underbrace{f(1)}_{f(1)}+\underbrace{(f(2)+f(2))}_{2\times f(2)}+\underbrace{(f(4)+f(4)+f(4)+f(4))}_{4\times f(4)}+\cdots\\ &\le\sum_{n=0}^\infty\;2^n\,f(2^n)\\ \end{align} The same way \begin{align} \sum_{n=0}^\infty\;2^n\,f(2^n)&= f(1)+(f(2)+f(2))+(f(4)+f(4)+f(4)+f(4))+(f(8)+\cdots\\ &\le(f(1)+f(1))+(f(2)+f(2)+f(3)+f(3))+(f(4)+f(4)+\cdots\\ &\le 2\sum_{k=1}^\infty\;f(k)\\ \end{align} so that $\;\sum_{k=1}^\infty\;f(k)\;$ will converge iff $\;\sum_{n=0}^\infty\;2^n\,f(2^n)$ converges.

Since $\,f(k):=\dfrac 1{k^{\,p}}\,$ satisfies the required criteria for any fixed non-negative power $p$ we deduce that $\;\sum_{k=1}^\infty\;\dfrac 1{k^{\,p}}\;$ will converge iff $\;\sum_{n=0}^\infty\;2^n\,\dfrac 1{2^{n\,p}}=\sum_{n=0}^\infty\;2^{n\,(1-p)}\;$ that is if and only if $\,p>1$.

The following illustration may clarify this. The picture at the middle shows the inequality in $(1)$ : the total area can't exceed the sum of the areas of all the $2^n$-width grey rectangles. In the specific case $p=1\,$ each of these rectangle will have an area of $1$ (the actual area can't differ from that by a factor larger than $2$) while for larger values of $p$ the areas will decrease exponentially and thus be finite. rectangles

The same reasoning applies for $\int \frac 1{x^p}\,dx$ from the integral test.

For alternative and possibly more convincing proofs of the divergence of the harmonic series see Kifowit's answers in "The Harmonic Series Diverges Again and Again" and "More Proofs of Divergence of the Harmonic Series".


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