Timeline for What is $\lim_{n\to\infty} \displaystyle \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k}\left(4^{-k}\binom{2k}{k}\right)^{\frac{2n}{\log_2{n}}}\,?$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 15, 2014 at 21:07 | vote | accept | Simd | ||
| Apr 15, 2014 at 6:51 | comment | added | S. Carnahan♦ | @Anush Yes. The OP's exponent is precisely what is needed to cancel the $n \log 2$ from $\log \binom{n}{2k}$ for optimal $k$. This is why I needed to consider the sub-leading term with the $\pi c$ computation. If the exponent is smaller, the sums diverge even more wildly, and if the exponent is larger, the sums converge to 1. | |
| Apr 15, 2014 at 6:28 | comment | added | Simd | Can I ask my question again in that case? Is the OP right that convergence depends on the constant in the exponent? | |
| Apr 15, 2014 at 5:00 | comment | added | S. Carnahan♦ | @GregMartin You're right, thanks. I've made quite a mess of this question. | |
| Apr 15, 2014 at 4:57 | history | edited | S. Carnahan♦ | CC BY-SA 3.0 |
More sign errors
|
| Apr 14, 2014 at 7:21 | comment | added | Greg Martin | Shouldn't the function in (4) be $n$ times $-2c\log2c - (1-2c)\log(1-2c)$ rather than $-2c\log2c + (1-2c)\log(1-2c)$? | |
| Apr 14, 2014 at 0:27 | history | edited | S. Carnahan♦ | CC BY-SA 3.0 |
Sign error in log binomial
|
| Apr 14, 2014 at 0:07 | history | edited | S. Carnahan♦ | CC BY-SA 3.0 |
More explicit estimates.
|
| Apr 13, 2014 at 14:11 | comment | added | S. Carnahan♦ | @Anush Yes. In the fourth expansion of my list, the optimal coefficient of $n$ is about $0.14676$, and the coefficient of $n$ from the fifth sum is $-\log 2$. Thus, if you divide the exponent by more than about 4.722967, the limit diverges. [Edit: this is not accurate - see later comments.] | |
| S Apr 13, 2014 at 11:39 | history | suggested | user44143 | CC BY-SA 3.0 |
improved formatting
|
| Apr 13, 2014 at 11:34 | review | Suggested edits | |||
| S Apr 13, 2014 at 11:39 | |||||
| Apr 13, 2014 at 10:51 | comment | added | Simd | Is the OP right that convergence depends on the constant in the exponent? I can't immediately tell from your answer. | |
| Apr 13, 2014 at 10:19 | history | edited | S. Carnahan♦ | CC BY-SA 3.0 |
Forgot a 1/2 in my derivative.
|
| Apr 13, 2014 at 9:44 | history | answered | S. Carnahan♦ | CC BY-SA 3.0 |