0

I find it hard to wrap my head around this.

Consider the following harmonic sum:

$$\sum_{k=1}^\infty \frac{1}{k},$$

in which the individual terms obviously converge to $0$ as $k\to\infty$, but the sum diverges.

Then in comparison consider the following geometric sum:

$$\sum_{k=1}^\infty r^k\quad\text{when}\quad\lvert r\rvert <1,$$

in which the individual terms converge to $0$, but apparently fast enough so the sum also converges.

When you plot the first one, it seems as if converging but kind of understandably it is not. Is there any clarifying intuitive ways to approach the difference between the two sums here? The difference seems so slight that it kind of makes the reality feel relative.

metamorphy
  • 39,111
Lassi
  • 13
  • What makes you say that the difference between the harmonic series and a geometric series is “slight”? – Martin R Dec 01 '22 at 15:56
  • Slight on the sense that the convergence-divergence dynamic seems almost the same – Lassi Dec 01 '22 at 15:58
  • 1
    i don't understand the similarities you are drawing between harmonic sum and geometric sum, but your question is nonetheless interesting. There is a particular "rate" at which the sums increase, and this sum must cross a particular threshold to diverge. I have always considered the harmonic sum to be right on the threshold, if not the threshold itself. but i have no proof of this – RyRy the Fly Guy Dec 01 '22 at 16:01
  • https://math.stackexchange.com/q/1128094/1104092

    I just found one topic which kind of touches on the idea, but focusing on the two different harmonic means. But yea, I am not a mathemathican so my brains might not be wired enough to differentiate between the two I am proposing here. Just to add, of course I understand the basic differences between geometric and harmonic series, but in this particular setting the dynamics seem similar

     – Lassi Dec 01 '22 at 16:06
  • 1
    @RyRytheFlyGuy $\sum \frac1{k^c}$ diverges iff $c\le 1$ so in that sense it is the threshold but in a sense $\sum \frac1{k\log(k)}$ is closer to the threshold and so it goes on. Similarly $\sum\limits_{p\text{ prime}} \frac1p$ – Henry Dec 01 '22 at 16:06
  • See also https://math.stackexchange.com/a/452074/42969: For every converging series there is another converging series with “much bigger” terms, and for every diverging series there is another diverging series with “much smaller” terms. – Which means that a “fastest growing converging series” or a “slowest growing diverging series” does not exist. – Martin R Dec 01 '22 at 16:24
  • 2
    The question is basically, "I can't accept that things are the way they are." Well maths has many results that seem strange at first. Get over it... But also the maths aspect of this is that, given any convergent sequence, you can find a "much larger" sequence that converges. Similarly, given any divergent sequence, you can find a "much smaller" sequence that diverges. There is a popular question about this on this site, but I can't find it right now. – Adam Rubinson Dec 01 '22 at 16:35
  • Thank you all. I suppose this was exactly what I needed. Different perspectives are invaluable. – Lassi Dec 01 '22 at 20:04
  • A much broader intuition is about what happens to a proposition when infinity is involved. When the result is not unique or very simple, then usually it could become complicated and very complicated. You might drill down infinitely. Infinity is a different kind of beast. – Apass.Jack Dec 01 '22 at 21:25

0 Answers0