0

I had a homework question asking me to evaluate the series $1 + \frac{1}{2} - 1 + \frac{1}{3} + \frac{1}{4} - \frac{1}{2}...$

Ultimately the solution was just to combine the negative terms with the ones right before them to get a series expansion for $\ln(2)$. However, while looking for the solution I noticed that after each negative term the sum was equal to $H_{2n} - H_n$, meaning $\lim \limits_{n \to \infty} H_{2n} - H_n = \ln(2)$. This made me curious: What about for $H_{3n}$, $H_{4n}$, etc? Generally, what does it approach for $H_{\lfloor xn \rfloor}$?

MilesZew
  • 735
  • Can you be more explicit in explaining exactly which series you are asking about? – Dan Asimov Mar 20 '24 at 15:46
  • @DanAsimov well the function I'm asking about is $f(x) = \lim \limits_{n \to \infty} H_{\lfloor xn \rfloor} - H_n$ but the series I used to explain where the question comes from is just a rearrangement of $1 - 1 + \frac{1}{2} - \frac{1}{2} ...$ where you add the next two positive terms, then the next negative term, repeat – MilesZew Mar 20 '24 at 15:54
  • Changing the order of an infinite series will change the outcome you get, please note that only finitely many terms of a series can be described with associativity. Theoretically you could rearrange the series to get any value. – Amy Skinner Mar 20 '24 at 15:57
  • 1
    Using $H_n = \log n+\gamma+O(1/n)$, we see that $H_{\lfloor xn\rfloor}-H_n = \log x + O(1/n)\to \log x$ as $n\to\infty$. – Riemann Mar 20 '24 at 16:16

0 Answers0