I've been trying to understand Oresme's proof that the harmonic series diverges since it's greater than the series of halves, which diverges.

I'm struggling to capture an aspect of the relationship which I think can be expressed as: "the series of halves is not surjective on the harmonic series".

My best idea so far is that the series of halves goes up to $\log_2$ of the $n$ for the harmonic series. Does that seem right?

$$ \begin{alignat*}{10} \sum_{k=1}^{n}\frac{1}{k} &=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\dots\\\\ &=\frac{1}{1}+\frac{1}{2}+[\frac{1}{3}+\frac{1}{4}]+[\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}]\dots\\\\ &> \begin{cases} \begin{aligned} &\frac{1}{1}+\frac{1}{2}+[\frac{1}{4}+\frac{1}{4}]+[\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}]\dots\\\\ =&1+\begin{cases} \begin{aligned} &\frac{1}{2}+\frac{2}{4}+\frac{4}{8}\dots\\\\ = &\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\dots\\\\ = &\sum_{k=1}^{\log_2(n)}\frac{1}{2}\\\\ = &\frac{\log_2(n)}{2}\\ = &\log_2(n)-1 \end{aligned} \end{cases}\\ \end{aligned} \end{cases} \end{alignat*}$$


1 Answer 1


Another way to write this.

Consider the first $m$ groups of $2^k$ terms for $k = 0$ to $m-1$. The $k$-th group is $g(k) =\sum_{j=2^k}^{2^{k+1}-1} \dfrac1{j} $.

Since $j$ for each term is less than $2^{k+1}$, and there are $2^k$ terms, the sum satisfies $g(k) \gt\sum_{j=2^k}^{2^{k+1}-1} \dfrac1{2^{k+1}} =\dfrac{2^k}{2^{k+1}} =\dfrac12$.

Therefore, the sum of $m$ of these groups is greater than $\dfrac{m}{2}$, and this can be made arbitrarily large.


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