Zulip Chat Archive

Stream: Is there code for X?

Topic: (n+1)^{1/n} tends to 1


Jiang Jiedong (Nov 04 2024 at 18:22):

Hi everyone,
Is this result:
limn(n+1)1/n=1\lim_{n \to \infty} (n+1)^{1/n} = 1
currently in Mathlib? If not, what theorems should I add to mathlib to get this result? Thank you!

Damiano Testa (Nov 04 2024 at 18:34):

I hope that sorry would be required somewhere...

Etienne Marion (Nov 04 2024 at 18:39):

This tends to 11

Jiang Jiedong (Nov 04 2024 at 18:41):

Etienne Marion said:

This tends to 11

Thank you! This is a stupid typo of me. Modified.

Bjørn Kjos-Hanssen (Nov 04 2024 at 18:59):

https://leanprover-community.github.io/mathlib4_docs/Mathlib/Analysis/Calculus/LHopital.html

Kevin Buzzard (Nov 04 2024 at 23:25):

I would take logs and then hope that we have log(n) is o(n).

Yan Yablonovskiy (Nov 04 2024 at 23:50):

Jiang Jiedong said:

Hi everyone,
Is this result:
limn(n+1)1/n=1\lim_{n \to \infty} (n+1)^{1/n} = 1
currently in Mathlib? If not, what theorems should I add to mathlib to get this result? Thank you!

Taking logs and justifying via regular limit laws and continuity.

Then as others have also said, can use LHopitals. A very oblique but curious proof could be using prime number theorem, if it were in mathlib.

Damiano Testa (Nov 04 2024 at 23:56):

There is docs#tendsto_rpow_div. You can then argue that your sequence is bounded below by n ^ (1 / n) and above by (2 * n) ^ (1 / n).

Jiang Jiedong (Nov 05 2024 at 09:23):

Damiano Testa said:

There is docs#tendsto_rpow_div. You can then argue that your sequence is bounded below by n ^ (1 / n) and above by (2 * n) ^ (1 / n).

Thanks! This seems to be the shortest path.


Last updated: May 02 2025 at 03:31 UTC