# Constant sequence of partial sums in a diverging series

In the harmonic series, we have $$|H_{2n}−H_n|\geq \frac{1}{2}$$ for all $$n$$, which implies divergence. However, the partial sums from $$n$$ to $$2n$$, evaluated at $$n$$, equal $$\ln(2)$$ for all $$n$$. Doesn't this imply the sequence of partial sums has converged to the value $$\ln(2)$$, which in turn, implies the series should converge? I feel like I'm not understanding something fundamental about the Cauchy criterion and convergence etc -- is this not a sequence of partial sums at all, due to the funny things we're doing with the interval? Thanks for your help.

• You have to be careful as to which partial sums you're talking about. Convergence of a series means that the partial sums from $0$ to $n$ converge as $n\to\infty$. You've shown that the partial sums from $n$ to $2n$ converge, and that's entirely different. – Andreas Blass Aug 15 '20 at 16:54
• @AndreasBlass So what I have shown is, in effect, that a sequence of partial sums converges, not the sequence? Could we refer to this as a convergent subseries then? – rage_man Aug 15 '20 at 22:25
• Not is "subseries" is intended to mean what I think it does. Even for a subseries, the relevant partial sums would all have to begin at the same place. – Andreas Blass Aug 15 '20 at 22:32

## 1 Answer

First, a minor thing: the partial sums from $$n$$ to $$2n$$ approach $$\ln{2}$$, but will never actually equal it. (Why?)

Second, more major thing: In fact, what you have shown is that the sequence of partial sums $$\{ H_n\}$$ is not Cauchy, and thus not convergent. Indeed, if it were Cauchy, then by definition $$|H_{2n} - H_n| \to 0$$. This is because for any $$\epsilon > 0$$, there would have to exist $$N(\epsilon)$$ for which $$|H_m - H_n| < \epsilon$$ whenever $$m, n > N(\epsilon)$$; we then choose $$m = 2n$$ here.

• For the first question, my guess is that's the definition of a limiting value -- it approaches but doesn't equal. Is this too hand - wavy? Thank you for the other feedback -- between Andrea's comment and yours it makes a lot more sense. – rage_man Aug 15 '20 at 20:20
• @rage_man: there is no requirement that the function never equals its limiting value. For example, constant functions for example equal their limiting value everywhere! Another less boring example: $\sin{x}/x \to 0$ as $x \to \infty$, and it attains this value at all multiples of $\pi$. The reason there is never equality is that all partial sums, and thus their differences, are rational, while $\ln{2}$ is irrational. – Paco Adajar Aug 15 '20 at 20:48
• Ha, wrong on two accounts then -- I wouldn't have thought of any of that. Thanks again! – rage_man Aug 15 '20 at 22:15