## Posts Tagged ‘binomial’

### Roots of unity and binomial sums

Thanks to Robert Israel who answered my question in mathstackexchange, we have a generalization of the binomial sums of the previous post.  Interestingly, it turns out roots of unity are involved.  Given,

$w = e^{2\pi\, {\rm i}/k}$

where k is an even integer then,

\begin{aligned}&\sum_{n=0}^\infty \frac{(2z)^{kn}}{\binom{kn}{kn/2}} = \frac{1}{1-z^k}+\frac{1}{k}\sum_{j=0}^{k-1}\frac{w^j z \arcsin(w^j z)}{(1-w^{2j}z^2)^{3/2}}\end{aligned}

for appropriate z such that the sum converges.  For the special case when,

$2z =w^{1/2}=e^{\pi\,{\rm i}/k}$

then,

\begin{aligned}&\sum_{n=0}^\infty\frac{(-1)^n}{\binom{kn}{kn/2}}=\frac{2^k}{2^k+1}+\frac{1}{k}\sum_{j=0}^{k-1}\frac{w^{j+1/2} \arcsin(\tfrac{1}{2}\,w^{j+1/2})}{2(1-\tfrac{1}{4}\,w^{2j+1})^{3/2}}\end{aligned}

Note that the terms are complex, but the sum is a real number so they must come in conjugate pairs. The arcsin of a complex root of unity can be given as,

$\arcsin(\frac{1}{2}\,e^{\pi\,{\rm i}/k}) = -\arcsin\big(\frac{-a+b}{4}\big)+{\rm i}\ln\Big(\frac{a+b+\sqrt{-6+2ab}}{4}\Big)$

where,

\begin{aligned}a &= \sqrt{5+4\cos(\pi/k)}\\ b &= \sqrt{5-4\cos(\pi/k)}\end{aligned}

With this transformation, it is now possible to have an expression all in real terms.  The case k = 2, 4 was given in the previous post.  For k = 6, we have the counterpart to Sprugnoli’s equality as,

\begin{aligned}\sum_{n=0}^\infty \frac{(-1)^n}{\binom{6n}{3n}}&=\tfrac{64}{65}-\tfrac{2\sqrt{26}\,(1+7\sqrt{13})}{3\cdot13^2\,\sqrt{1+\sqrt{13}}}\arcsin\left(\tfrac{-\sqrt{6}}{2\sqrt{5+\sqrt{13}}}\right)\\&-\tfrac{\sqrt{26}\,(-1+7\sqrt{13})}{3\cdot13^2\,\sqrt{-1+\sqrt{13}}}\ln\left(\tfrac{\sqrt{7+2\sqrt{13}}\,+1}{\sqrt{7+2\sqrt{13}}\,-1}\right)-\tfrac{4\sqrt{5}}{3\cdot5^2}\ln\left(\tfrac{1+\sqrt{5}}{2}\right) = 0.95106\dots\end{aligned}

Note that the prime factors of 65 are 5 and 13, and the square root of both appear above. However, for k = 8, while the expression contains the fraction $\frac{256}{257}$ as expected, the argument of the log and arcsin do not factor over the quadratic extension $\sqrt{257}$, but rather only over $\sqrt{2}$.  Furthermore, the argument of the log for both k = 6, 8 are no longer simply expressible in terms of the Dedekind eta function, so observations for lower k do not generalize to higher ones.

### Fermat primes and Binomial sums

We have,

\begin{aligned} \sum_{n=0}^\infty \frac{(-1)^n}{\binom n{n/2}} &= \frac{4}{3}-\frac{4\pi\sqrt{3}}{27}\\[2.5mm] \sum_{n=0}^\infty \frac{(-1)^n}{\binom {2n}n} &= \frac{4}{5} - \frac{4\sqrt{5}}{25}\ln\left(\frac{1+\sqrt{5}}{2}\right) \end{aligned}

For the next step, Renzo Sprugnoli gave the Ramanujan-like identity,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}+\frac{4\sqrt{34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\arctan\left(\frac{\sqrt{2}}{\sqrt{-1+\sqrt{17}}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) = 0.846609\dots \end{aligned}

(The sign of the third term has been changed by this author.)  However, to make it more symmetrical, we can express the arctan in terms of the log function.  Since,

\begin{aligned}&\arctan(z) = \frac{i}{2}\ln\left(\frac{1-i z}{1+i z}\right)\end{aligned}

then,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{2\sqrt{-34}\,(-2+\sqrt{17}\,)}{17^2\,\sqrt{-1+\sqrt{17}}}\ln\left(\frac{\sqrt{-1+\sqrt{17}}+\sqrt{-2}}{\sqrt{-1+\sqrt{17}}-\sqrt{-2}}\right)\\&-\frac{2\sqrt{34}\,(2+\sqrt{17}\,)}{17^2\,\sqrt{1+\sqrt{17}}} \ln\left(\frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\right) =0.846609\dots \end{aligned}

In this manner, it reduces to the concise,

\begin{aligned}\sum_{n=0}^\infty\frac{(-1)^n}{\binom{4n}{2n}}&=\frac{16}{17}-\frac{x_1}{17}\ln(y_1)-\frac{x_2}{17}\ln(y_2)\end{aligned}

where, $x_1,x_2$ and $y_1,y_2$ are the appropriate roots of,

\begin{aligned} &289x^4-799x^2-676 = 0\\ &y^4-5y^3+4y^2-5y+1 = 0\end{aligned}

I found that, curiously, the argument of the log can be expressed in terms of the Dedekind eta function, $\eta(z)$.  Let,

\begin{aligned} t_1 &=\frac{1+\sqrt{-5}}{2}\\ t_2 &= \frac{1+\sqrt{-17}}{2}\\ \zeta_{48} &=\exp(\pi i/24)\end{aligned}

then,

\begin{aligned} &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_1)}{\eta(2t_1)}\right)^4 = \frac{1+\sqrt{5}}{2}\\ &\frac{1}{2}\left(\frac{\zeta_{48}\, \eta(t_2)}{\eta(2t_2)}\right)^4 = \frac{\sqrt{1+\sqrt{17}}+\sqrt{2}}{\sqrt{1+\sqrt{17}}-\sqrt{2}}\end{aligned}

Is this coincidence?  Furthermore, using these as the argument of the polylogarithm,

\begin{aligned} &L_s(z) = \text{Li}_s (z) = \sum_{k=1}^\infty\frac{z^k}{k^s}\end{aligned}

one can find a polylogarithm ladder to express Apery’s constant.  For example, getting the square root and reciprocal of $y_2$ so that z < 1,

\begin{aligned} z &= \sqrt{\frac{\sqrt{1+\sqrt{17}}-\sqrt{2}}{\sqrt{1+\sqrt{17}}+\sqrt{2}}} = 0.480533\dots\end{aligned}

then,

$-12L_3(z)+75L_3(z^2)-68L_3(z^3)-33L_3(z^4)+43L_3(z^6)+12L_3(z^8)-7L_3(z^{12})+2\log^3(1/z) = 3\zeta(3)$

A simpler one exists for the other argument. The next step, of course, is,

\begin{aligned}&\sum_{n=0}^\infty \frac{(-1)^n}{\binom{8n}{2n}}=\,?\end{aligned}

Since the first three Fermat primes 3, 5, 17 have already appeared, it should be interesting to conjecture if 257 will be next.

### A missing binomial sum identity?

D.Bailey, J. Borwein, and D.Bradley found the beautiful pair involving binomial sums.  In Theorem 1 of this paper (2008), let x $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2-x^2} &= 3\,\sum_{k=1}^\infty \frac{1}{k^2 \binom{2k}k (1-x^2/k^2)} \prod_{m=1}^{k-1} \left(\frac{1-4x^2/m^2}{1-x^2/m^2}\right)\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^3(1-x^4/k^4)} &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3 \binom{2k}k (1-x^4/k^4)} \prod_{m=1}^{k-1} \left(\frac{1+4x^4/m^4}{1-x^4/m^4}\right)\end{aligned}

When x = 0, they reduce into,

\begin{aligned} \zeta(2) &= 3\,\sum_{k=1}^\infty \frac{1}{k^2\binom{2k}k }\\ \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\binom{2k}k }\end{aligned}

However, there is a third single-term equality,

\begin{aligned} \frac{17}{36}\,\zeta(4) &= \sum_{k=1}^\infty \frac{1}{k^4\binom{2k}k }\end{aligned}

so there might be a third identity that reduces to this as the special case x = 0.

To compare, there are three identities such that as $x \to 0$, then those zeta values are the respective limit.  For x $\not=$ integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{3k^2+x^2}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)-1}{x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom{2k}k}\,\frac{5k^2-x^2}{2(k^2-x^2)}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{-\psi^{(0)}(1-x)-\psi^{(0)}(1+x)-2\gamma}{2x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{1}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)+3\cos(\pi x/3)-4}{4x^4}\end{aligned}

The first two were found by Leshchiner and Koecher, respectively, while the third is Theorem 2 in the same paper by Bailey, Borwein, and Bradley. The function $\psi^{(0)}$ is given in Mathematica as,

$\psi^{(0)}(z) = \text{PolyGamma[0,z]}$

while $\gamma$ is the Euler-Mascheroni constant.  So are the Bailey-Borwein-Bradley pair of binomial sum identities in fact a triplet?

### Apery-like formulas for zeta(2n)

It is well-known that,

\begin{aligned}\zeta(2) &= 3\sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\end{aligned}

D. Bailey, J. Borwein, D. Bradley gave a generalization. First define,

\begin{aligned}&A(a_0) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k}\\ &A(a_0, a_1, a_2,\dots) = \sum_{k=1}^\infty \frac{1}{k^{a_0}\,\binom{2k}k} \sum_{p=1}^{k-1}\frac{1}{p^{a_1}} \sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\dots\end{aligned}

Obviously,

$\zeta(2) = 3A(2)$

However, a little experiment with Mathematica’s LatticeReduce command will show there are two solutions for $\zeta(4)$,

\begin{aligned} &a\big(\zeta(4)-3A(4)+9A(2,2)\big) =0\\ &b\big(5\zeta(4)-10A(4)-6A(2,2)\big) = 0\end{aligned}

where {a, b} are scaling variables.  Adding the two together,

$(a-5b)\zeta(4)-(3a-10b)A(4)+3(3a+2b)A(2,2) = 0$

hence there are an infinite number of solutions.  For appropriately chosen {a,b}, we can also eliminate one term. Thus,

\begin{aligned} \zeta(4) &= \frac{36}{17}A(4)\\ &=\frac{108}{5}A(2,2)\end{aligned}

For $\zeta(6)$, there are now three solutions. Given the five terms,

$\zeta(6),\, A(6),\, A(4,2),\, A(2,4),\, A(2,2,2)$

then the coefficients such that their sum is equal to zero are,

\begin{aligned} &\text{1st sol:}\; (5, -9, 1, -15, 3)\\ &\text{2nd sol:}\; (2, -7, 23, 9, 15)\\ &\text{3rd sol:}\;\, (10,-17,-3,6,-36)\end{aligned}

Using the same approach above, we can eliminate two of the terms.  One solution has an interesting number pop up,

$163\zeta(6) = 288A(6)+432A(2,4)$

though the appearance of 163 is probably only a coincidence.   See Bailey, Borwein, Bradley’s paper for more details.

### Borwein and Bradley’s Apery-like formulas for zeta(4n+3)

Apery gave,

\begin{aligned} \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom {2k}k}\end{aligned}

J. Borwein and D. Bradley found this can be generalized to $\zeta(4n+3)$. Define the functions,

\begin{aligned} &B(a_0)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\\ &B(a_0,a_1,a_2,\dots)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\; \sum_{p=1}^{k-1} \frac{1}{p^{a_1}}\;\sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\;\dots \end{aligned}

then,

\begin{aligned} \frac{2}{5}\,\zeta(3) &= B(3)\\ \frac{2}{5}\,\zeta(7) &= B(7)+5B(3,4)\\ \frac{2}{5}\,\zeta(11) &= B(11)+5B(7,4)-\frac{15}{2}B(3,8)+\frac{25}{2}B(3,4,4)\\ \frac{2}{5}\,\zeta(15) &= B(15)+5B(11,4)-\frac{15}{2}B(7,8) +\frac{25}{2}B(7,4,4)+\frac{130}{6}B(3,12)\\&-\frac{225}{6}B(3,8,4)+ \frac{125}{6}B(3,4,4,4) \end{aligned}

and so on.  Beautiful, aren’t they? Notice that all the $a_i$ (excepting $a_0$) are all divisible by 4. This infinite family has a generating function. Let z $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^3(1-z^4/k^4)}&=\frac{5}{2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^3\;\binom {2k}k} \frac{1}{1-z^4/k^4}\prod_{j=1}^{k-1}\frac{1+4z^4/j^4}{1-z^4/j^4}\end{aligned}

On the other hand, for s = 4n+1,

\begin{aligned} 2\,\zeta(5) &= 4B(5)-5B(3,2)\\[2.5mm] 4\,\zeta(9) &= 9B(9)-5B(7,2)+20B(5,4)+45B(3,6)-25B(3,4,2)\\[2.5mm] 12\,\zeta(13) &= 28B(13)-10B(11,2)+90B(9,4)\\&+90B(7,6)-50B(7,4,2)-60B(5,8)+100B(5,4,4)\\&-310B(3,10)+75B(3,8,2)+450B(3,6,4)-125B(3,4,4,2)\end{aligned}

with this version for $\zeta(13)$ found by Jim Cullen.  There are various versions for both s = 4n+1 and 4n+3.  For example, for $\zeta(7)$, we have the relations,

\begin{aligned} 4\zeta(7) &= 8B(7,0)\,-\,5B(3,4)\,-\,8B(5,2)+5B(3,2,2)\\ 0 &= -2B(7,0) - 55B(3,4)-8B(5,2)+5B(3,2,2)\end{aligned}

Eliminating the last two terms will yield the shorter relation given by Borwein and Bradley. There is a generating function for all s = 2n+1, but none is known that is only for s = 4n+1. See Apery-Like Formulae for $\zeta(4n+3)$ for more details.

### Binomial sums and the zeta function

Recall the three sequences,

$C_n = \sum_{k=0}^n {\binom n k} {\binom {n+k}k} = 1, 3, 13, 63, 321, 1683\dots$

$B_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k} = 1, 3, 19, 147, 1251, 11253\dots$

$A_n = \sum_{k=0}^n {\binom n k}^2 {\binom {n+k}k}^2 = 1, 5, 73, 1445, 33001, 819005\dots$

Equivalently,

$C_n = \,_2F_1(-n,n+1;\,1;\,-1)$

$B_n = \,_3F_2(-n,-n,n+1;\,1,1;\,1)$

$A_n = \,_4F_3(-n,-n,n+1,n+1;\,1,1,1;\,1)$

where $_pF_q$ is the generalized hypergeometric function.  Then it is known that,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\, C_n C_{n+1}} = \frac{1}{2}\log(2)\\ &\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2 B_n B_{n+1}} = \frac{1}{5}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3 A_n A_{n+1}} = \frac{1}{6}\, \zeta(3)\end{aligned}

Beautiful, aren’t they? Since the numbers increase at a near-geometric rate (for example, $\frac{A_{n+1}}{A_n} \to (1+\sqrt{2}\,)^4 \approx 33.97$  as n  goes to infinity), then the convergence is very fast.

We also have the nice evaluations,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{n\,\binom {2n}n} = \frac{\pi}{3\sqrt{3}}\\ &\sum_{n=1}^\infty \frac{1}{n^2\,\binom {2n}n} = \frac{1}{3}\,\zeta(2)\\ &\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} = -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right) \\ &\sum_{n=1}^\infty \frac{1}{n^4\,\binom {2n}n} = \frac{17}{36}\,\zeta(4)\\ &\sum_{n=1}^\infty \frac{1}{n^5\,\binom {2n}n} = -\frac{19}{3}\,\zeta(5) +\frac{2}{3}\,\zeta(2)\zeta(3)+\frac{\pi\sqrt{3}}{2^3\cdot 3^2}\left(\zeta(4, \tfrac{1}{3})-\zeta(4,\tfrac{2}{3}) \right)\\ &\sum_{n=1}^\infty \frac{1}{n^6\,\binom{2n}n} = \;\;?\\ &\sum_{n=1}^\infty \frac{1}{n^7\,\binom{2n}n} = -\frac{493}{24}\zeta(7)+2\zeta(2)\zeta(5)+\frac{17}{18}\zeta(3)\zeta(4)+\frac{11\pi\sqrt{3}}{2^5\cdot 3^4}\left(\zeta(6,\tfrac{1}{3})-\zeta(6,\tfrac{2}{3})\right)\\ \end{aligned}

with the Riemann zeta function $\zeta(s)$ and the more general Hurwitz zeta function $\zeta(s,a)$,

\begin{aligned} &\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\\&\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^s} \end{aligned}

respectively.  (Note that for $a = 1$, the Hurwitz reduces into the Riemann.)  The expression for p = 5 in the paper here used Dirichlet L-functions, but a poster from mathstackexchange gave it in terms of the Hurwitz zeta.  The one for p = 7 is from Mathworld’s article on central binomial coefficients.

However, none are known for p > 7 (as well as p = 6).  Based on odd p, it is easy to assume that the next has the form,

\begin{aligned} & a_0\sum_{n=1}^\infty \frac{1}{n^9\,\binom {2n}n} = a_1\,\zeta(9) +a_2\,\zeta(2)\zeta(7)+a_3\, \zeta(3)\zeta(6)+a_4\, \zeta(4)\zeta(5)+a_5\,\pi\sqrt{3}\,H_8\\ \end{aligned}

where,

$H_8 = \left(\zeta(8,\tfrac{1}{3})-\zeta(8,\tfrac{2}{3})\right)$

and the six $a_i$ are integers.  One can use Mathematica’s LatticeReduce function (which employs an integer relations algorithm) to find them, if any exists. Unfortunately, it didn’t find any exact relation, nor for analogous forms for prime p = 11 or 13.  Either my old version of Mathematica is just not powerful enough, or odd p > 7 do not have analogous forms to the ones above.

Can you find the next in the family?