Timeline for Does any method of summing divergent series work on the harmonic series?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2019 at 4:07 | comment | added | user76284 | The limit is approaching 1 from the left, so the actual value at $x=1$ (which does diverge) doesn't matter. What matters are the values "on the way there" or "on that path", i.e. for $x < 1$, all of which go to 0 pointwise. The limit itself, so to speak, doesn't know anything or care about the "actual value" at $x=1$, only what things look like as it approaches $x=1$. Incidentally this is why the order of limits is important. Let me know if this makes sense. | |
| Jul 6, 2019 at 3:59 | comment | added | j0equ1nn | But isn't $x$ approaching $1$, rather than $0$? | |
| Jul 3, 2019 at 0:40 | comment | added | user76284 | @j0equ1nn $\ln n$ indeed diverges, but $x^n$ goes to 0 faster on $[0, 1)$: desmos.com/calculator/eu9fpspoye | |
| Jul 2, 2019 at 23:45 | comment | added | j0equ1nn | That's a clever approach to this, and interesting to see the effect of switching the limits. One thing I'm not getting though: doesn't $\lim_{x\nearrow1}(\lim_{n\rightarrow\infty}(x^n\ln n))$ diverge due to $\ln n$ diverging? You have it as though that equals $0$. | |
| Jun 29, 2019 at 3:11 | history | answered | user76284 | CC BY-SA 4.0 |