Timeline for Limit of a sum with binomial coefficients
Current License: CC BY-SA 4.0
11 events
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| Mar 29, 2022 at 22:01 | comment | added | Max Alekseyev | @IosifPinelis: Yes, there should not be factor $k$ in the denominator (fixed now!). Thanks for noticing. To switch to $f(x)$, I use Lagrange–Bürmann formula. | |
| Mar 29, 2022 at 21:59 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Mar 29, 2022 at 17:51 | comment | added | Iosif Pinelis | @MaxAlekseyev : You wrote: "It follows that $S_k$ equals the coefficient of $x^k$ in $$\frac{1+2x}{2(1+x)(1-x)^kk\binom{2k-1}{k}}.$$" This does not seem to be true, even for $k=2$. A typo there? It is also unclear how you go from there to $f(x)$. | |
| Mar 29, 2022 at 17:01 | comment | added | Iosif Pinelis | @MaxAlekseyev : Thank you for your response. | |
| Mar 28, 2022 at 13:49 | comment | added | Max Alekseyev | @IosifPinelis: The function ${\cal S}(x) - \frac{2}{3}\frac{1}{1-x}$ converges for $|x|<8$ and thus its coefficient of $x^k$ is smaller than $(\frac18+\epsilon)^k$ giving the desired limit. | |
| Mar 28, 2022 at 4:31 | comment | added | Iosif Pinelis | I am not sure how you got the limit $2/3$ of $S_k$. Of course, by the Hardy–Littlewood tauberian theorem (en.wikipedia.org/wiki/…), you can get $S_1+\cdots+S_k\sim2k/3$. However, if e.g. $s_k=2/3+\sum_{j\ge1}1(k=j^2)$, then you get $\lim_{x\uparrow1}(1-x)\sum_{k\ge1}s_k x^k=2/3$, whereas $s_k$ does not converge to a limit. | |
| Mar 27, 2022 at 22:53 | comment | added | Max Alekseyev | This can probably be obtained from accurate asymptotic analysis of the coefficients of $(1-x){\cal S}(x)$. | |
| Mar 27, 2022 at 22:08 | comment | added | macat | I need some time to digest this magical solution. Could this approach also explain why $S_k$ is decreasing? | |
| Mar 27, 2022 at 21:19 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Mar 27, 2022 at 21:13 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
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| Mar 27, 2022 at 21:08 | history | answered | Max Alekseyev | CC BY-SA 4.0 |