Timeline for Limit of a sum with binomial coefficients
Current License: CC BY-SA 4.0
19 events
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| Mar 31, 2022 at 17:58 | vote | accept | macat | ||
| Mar 28, 2022 at 23:13 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 17:34 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 16:15 | comment | added | Jorge Zuniga | The method is named Creative Telescoping. But I guess it is tough to apply it pen-and-paper in this case. Expressions are not simple. Take a look, for instance, at the generating function obtained by Max Alekseyev. | |
| Mar 28, 2022 at 16:00 | comment | added | macat | If you did not observe the simple formula for $S_k$ (which enabled you to use Zeilberger's algorithm), then would there be still a method to obtain the recursion? | |
| Mar 28, 2022 at 15:49 | comment | added | Jorge Zuniga | This is a very old question. The answer is yes for linear recurrences. Even the asymptotic expansion can be obtained, Birkhoff and Trjitzinsky (1932). A good Introduction is found in Appendix B of Jet Wimp - Computation with Recurrence Relations (Applicable Mathematics Series) (1984), | |
| Mar 28, 2022 at 10:09 | comment | added | macat | Do you think that the recursion for $S_k$ could be used to prove the limit of the sequence? | |
| Mar 28, 2022 at 8:21 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 7:46 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 6:29 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 4:03 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 3:07 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 28, 2022 at 2:07 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 27, 2022 at 19:50 | comment | added | Jorge Zuniga | @macat From this simpler ratio you can apply Zeilberger´s algorithm to find the $\ell$-th order recurrence satisfied by $S_n$, as $$S_{n+1}=q_nS_n+q_{n-1}S_{n-1}+...q_{n-\ell}S_{n-\ell}$$. By analyzing the coefficients $q_n$ it should be possible to prove that $S_{n+1}<S_n$. In fact you should get $|q_j|<1,\ \ j=n-\ell, ..., n$ for $\ n>n_0$. | |
| Mar 27, 2022 at 17:06 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 27, 2022 at 16:37 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
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| Mar 27, 2022 at 11:34 | comment | added | macat | Update: This will not imply that $S_k$ is decreasing, because the error terms hidden by $O(1/n)$ may change signs (based on experiments). So, another reasoning is necessary to prove that $S_k$ is decreasing from $k=3$. | |
| Mar 27, 2022 at 1:12 | comment | added | macat | Thank you, this essentially answers why the limit of $S_k$ is $2/3$. I can not yet see the pen-and-paper proof, but it should be doable based on the cited paper. However, it is not clear to me if this also helps to prove that $S_k$ is decreasing from $k=3$. (What I need is that $S_k\geq 2/3$ for all $k$.) | |
| Mar 26, 2022 at 21:16 | history | answered | Jorge Zuniga | CC BY-SA 4.0 |