Timeline for Speed of convergence of $\zeta(2k)\to 1$?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2021 at 17:04 | comment | added | Richard Stanley | Bernoulli numbers have the well-known asymptotics $|B_{2n}|\sim 4\sqrt{\pi n}\left( \frac{n}{\pi e}\right)^{2n}$ as $n\to\infty$. | |
| Dec 16, 2021 at 11:14 | history | edited | Iceman | CC BY-SA 4.0 |
deleted 1 character in body
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| S Dec 16, 2021 at 9:55 | history | suggested | Buzz | CC BY-SA 4.0 |
fixed typos
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| Dec 16, 2021 at 0:33 | comment | added | Gerald Edgar | @LoïcTeyssier ... but how well can we estimate the asymptotic size of the Bernoulli numbers? | |
| Dec 16, 2021 at 0:26 | review | Suggested edits | |||
| S Dec 16, 2021 at 9:55 | |||||
| Dec 15, 2021 at 22:34 | vote | accept | Iceman | ||
| Dec 15, 2021 at 21:23 | comment | added | Loïc Teyssier | As you probably know, one can calculate $\zeta(2k)$ explicitly for integer $k$. | |
| Dec 15, 2021 at 21:19 | answer | added | Sean Eberhard | timeline score: 13 | |
| Dec 15, 2021 at 21:05 | comment | added | user334725 | You have $\zeta(k)=1+1/2^k+O(1/3^k)$ as $k\rightarrow\infty$. | |
| S Dec 15, 2021 at 20:41 | review | First questions | |||
| Dec 15, 2021 at 21:39 | |||||
| S Dec 15, 2021 at 20:41 | history | asked | Iceman | CC BY-SA 4.0 |