Timeline for Limit of the average of telescopic products
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2022 at 16:56 | vote | accept | giti | ||
| Aug 15, 2022 at 5:11 | comment | added | Sudix | I guess Mathematica used the identity $$ \sum_{k=a}^{b}\frac{\left(m-k\right)!}{\left(n-k\right)!}\ =\frac{\left(m-a+1\right)!}{\left(m-n+1\right)\left(n-a\right)!}-\frac{\left(m-b\right)!}{\left(m-n+1\right)\left(n-b-1\right)!} $$ together with the Stirling-approximation, i.e. substituting $ n!\leftarrow \sqrt{2\pi n}\;\left(\frac{n}{\mathrm{e}}\right)^n$ within the limit, then matched the products that converged and extracted them from the limit. The result is nowhere as clean as the given answer though | |
| Aug 14, 2022 at 20:49 | history | became hot network question | |||
| Aug 14, 2022 at 16:15 | answer | added | T. Amdeberhan | timeline score: 8 | |
| Aug 14, 2022 at 13:52 | comment | added | giti | The expression is simplified now. | |
| Aug 14, 2022 at 13:51 | history | edited | giti | CC BY-SA 4.0 |
deleted 3 characters in body
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| Aug 14, 2022 at 13:15 | comment | added | mathworker21 | just to make sure there's no typo, did you indeed mean to write $(k-1)(k-1)^k$ in the numerator of the RHS (instead of $(k-1)^{k+1}$)? | |
| Aug 14, 2022 at 13:02 | comment | added | giti | I tried it, but it could not print the steps. | |
| Aug 14, 2022 at 13:00 | comment | added | Barry Carter | Have you tried Mathematica's Trace mode? Might help | |
| S Aug 14, 2022 at 12:47 | review | First questions | |||
| Aug 14, 2022 at 17:08 | |||||
| S Aug 14, 2022 at 12:47 | history | asked | giti | CC BY-SA 4.0 |