14
$\begingroup$

Seven years ago, I asked about closed-forms for the binomial sum $$\sum_{n=1}^\infty \frac{1}{n^k\,\binom {2n}n}$$

Some alternative results have been made. Up to a certain $k$, it seems it can be expressed surprisingly by a log sine integral,

$$\rm{Ls}_n\Big(\frac{\pi}3\Big) = \int_0^{\pi/3}\Big(\ln\big(2\sin\tfrac{\theta}{2}\big)\Big)^{n-1}\,d\theta$$

and zeta function $\zeta(s)$. Hence,

$$\begin{aligned} \frac\pi2\,\rm{Ls}_1\Big(\frac{\pi}3\Big) &=\;3\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} =\zeta(2) \\ \frac\pi2\,\rm{Ls}_2\Big(\frac{\pi}3\Big) &=-\frac34\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} -\zeta(3) =-\frac\pi2\,\rm{Cl}_2\Big(\frac\pi3\Big)\\ \frac{6\pi}{35}\,\rm{Ls}_3\Big(\frac{\pi}3\Big) &=\frac{36}{17}\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} =\zeta(4)\\ \frac{2^3\pi}{3!}\rm{Ls}_4\Big(\frac{\pi}3\Big) &=-3\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} -19\zeta(5)-2\zeta(2)\zeta(3) \\ 32\pi\,\rm{Ls}_5\Big(\frac{\pi}3\Big) &=144 \sum_{n=1}^\infty \frac{1}{n^6\,\binom {2n}n} +2029\zeta(6)+192\zeta(3)^2 \\ \frac{2^8\pi}{5!}\rm{Ls}_6\Big(\frac{\pi}3\Big) &=-24 \sum_{n=1}^\infty \frac{1}{n^7\,\binom {2n}n} -493\zeta(7)-48\zeta(2)\zeta(5)-164\zeta(3)\zeta(4) \\ \end{aligned}$$

where $\rm{Cl}_2\big(\tfrac\pi3\big)$ is Gieseking's constant and other $\rm{Ls}_{2n}\big(\tfrac\pi3\big)$ can be found here. I found these using Mathematica's integer relations sub-routine. Unfortunately, either the pattern stops at this point, or some other variables are involved. Note that Borwein and Straub also found,

$$\pi\,\rm{Ls}_7\Big(\frac{\pi}3\Big) =-135\pi\,\rm{Gl}_{6,1}\Big(\frac{\pi}{3}\Big)+\Big(2152-\tfrac{103}{864}\Big)\zeta(8)+45\zeta(2)\zeta(3)^2\quad$$

where,

$$\rm{Gl}_{m,1}\Big(\frac{\pi}3\Big) = \sum_{n=1}^\infty \frac{\sum_{k=1}^{n-1}\frac1k}{n^m}\sin\Big(\frac{n\,\pi}3\Big)= \sum_{n=1}^\infty \frac{H_{n-1}}{n^m}\sin\Big(\frac{n\,\pi}3\Big)$$

with harmonic number $\rm{H}_n$.

Q: Can we bring this table higher and find a relation between the log sine integral $\rm{Ls}_7\big(\frac{\pi}3\big)$ and binomial sums?


$\color{blue}{Update:}$ Given the generalized log sine integral,

$$\rm{Ls}_m^{(k)}(\sigma) = \int_0^{\sigma}x^k\Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{m-1-k}\,dx$$

where the post was just the case $k=0$. If we use $k=1$ instead,

$$\rm{Ls}_m^{(1)}(\sigma) = \int_0^{\sigma} x\,\Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{m-2}\,dx$$

this paper mentions that Borwein et al found,

$$\sum_{n=1}^\infty \frac{1}{n^m\,\binom {2n}n} = \frac{(-2)^{\color{red}{m-2}}}{(m-2)!}\int_0^{\pi/3} x\,\Big(\ln\big(2\sin\tfrac{x}{2}\big)\Big)^{m-2}\rm{dx}$$

Note: The paper made a typo. (Corrected in red.)

$\endgroup$
  • $\begingroup$ The last formula (attributed here to Borwein et al) can't be correct. For $m=2$ is gives zero. $\endgroup$ – Dr. Wolfgang Hintze May 5 at 16:57
  • $\begingroup$ @ Tito Piezas III I meant m=1. But I repeat that the formula is wrong. Numerically, for m=1..4 we obtain from the formula the values {0., -1.09662, -1.04589, -1.02219} while the sum gives {0.6046, 0.548311, 0.522946, 0.511097}. $\endgroup$ – Dr. Wolfgang Hintze May 5 at 17:17
  • $\begingroup$ Using the usual procedure of replacing the sum by an integral I obtain the following expression for the sum $s(k) = \text{Li}_k\left(\frac{1}{4}\right)+\frac{1}{\Gamma(k)} \int_0^{\frac{\pi }{6}} \frac{2^k u \log ^{k-1}\left(\frac{1}{2 \sin (u)}\right)}{\cos ^2(u)} \, du$. The formula is correct numerically for the first few values of $k$. $\endgroup$ – Dr. Wolfgang Hintze May 5 at 17:22
  • $\begingroup$ @Dr.WolfgangHintze: I checked it. The authors made a small typo. The factor should be $$\frac{(-2)^{m-2}}{(m-2)!}$$ It now works. $\endgroup$ – Tito Piezas III May 5 at 17:28
  • 3
    $\begingroup$ Except for m=1 it is okay now. $\endgroup$ – Dr. Wolfgang Hintze May 5 at 17:47
3
$\begingroup$

Persistence pays off! Given the log sine integral,

$$\rm{Ls}_n\Big(\frac{\pi}3\Big) = \int_0^{\pi/3}\Big(\ln\big(2\sin\tfrac{\theta}{2}\big)\Big)^{n-1}\,d\theta$$

The case $\rm{Ls}_7\big(\frac{\pi}3\big)$ was elusive, but $\rm{Ls}_\color{red}8\big(\frac{\pi}3\big)$ was found. Hence,

$$\frac{2^{10}\cdot9\pi}{7!}\rm{Ls}_\color{red}8\Big(\frac{\pi}3\Big)+6^3 \sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n}\\=-13921\zeta(9)-6^4\zeta(2)\zeta(7)-6087\zeta(3)\zeta(6)-4428\zeta(4)\zeta(5)-192\zeta^3(3) $$

though I don't know why the lower level is elusive, or if higher levels can be found.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.