Timeline for Limit of a sum with binomial coefficients
Current License: CC BY-SA 4.0
9 events
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| Apr 3, 2022 at 13:06 | comment | added | macat | Let $Q_k=S_k-S_{k-1}$. By Zeilberger's algorithm we have that $Q_{k+2}=\frac{(7k-6)Q_{k+1}+kQ_{k}}{8k+12}$. We can compute $S_3,S_4$ and $S_5$, and verify that $Q_4>0$ and $Q_4\geq Q_5$. By induction, one gets that $Q_{k+2}=\frac{(7k-6)Q_{k+1}+kQ_{k}}{8k+12}\geq\frac{8k-6}{8k+12}Q_{k+1}$, hence $S_k$ is decreasing from $k=3$. | |
| Mar 30, 2022 at 23:03 | comment | added | macat | The main motivation behind this is an infinite sequence of similar questions (which are not the subjects of this thread), and the present one is the simplest of them. It does not make any difficulty to compute a finite prefix of the $P_k$'s in the current setting, but it is an issue if you need to do this infinitely many times. It would be also nicer to avoid these straightforward but quite lengthy computations. | |
| Mar 30, 2022 at 20:47 | comment | added | Iosif Pinelis | @macat : I think it might be possible to avoid the calculation for some of these $k$, but hardly for all of them -- at least with this method. But why do you want to avoid these straightforward calculations? | |
| Mar 30, 2022 at 19:58 | comment | added | macat | I have been trying to avoid the computation of $P_k$ for $k=4,\dots,9$. Do you think this is possible? | |
| Mar 28, 2022 at 7:19 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 28, 2022 at 6:39 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 28, 2022 at 6:39 | comment | added | Iosif Pinelis | Sorry, I do not know why formula (1) does not compile correctly after I click on the link. | |
| Mar 28, 2022 at 6:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 28, 2022 at 6:22 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |