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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.

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    $\begingroup$ 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. $\endgroup$ – Andreas Blass Aug 15 '20 at 16:54
  • $\begingroup$ @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? $\endgroup$ – rage_man Aug 15 '20 at 22:25
  • $\begingroup$ 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. $\endgroup$ – Andreas Blass Aug 15 '20 at 22:32
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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.

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  • $\begingroup$ 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. $\endgroup$ – rage_man Aug 15 '20 at 20:20
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    $\begingroup$ @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. $\endgroup$ – Paco Adajar Aug 15 '20 at 20:48
  • $\begingroup$ Ha, wrong on two accounts then -- I wouldn't have thought of any of that. Thanks again! $\endgroup$ – rage_man Aug 15 '20 at 22:15

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