5
$\begingroup$

If $k=1$, we have \begin{align} \sum_{n\ge1}\frac{1}{\binom{2n}{n}}&=\frac12\sum_{n\ge1}n\mathrm{B}(n,n)\\ &=\frac12\int _{0}^{1}\sum_{n\ge1}n(t-t^2)^{{n-1}}{\mathrm d}t\\ &=\frac12\int _{0}^{1}\frac{1}{(t-t^2-1)^2}{\mathrm d}t\\ &=\frac13+\frac{2\sqrt3\pi}{27} \end{align} I wonder if there exists a general formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$ with $k\in\Bbb N^*$. This may involve some ${}_pF_q$ functions.

$\endgroup$
6
$\begingroup$

We have $$ \binom{2n}{n}^{-1} = \frac{\Gamma(n+1)^2}{\Gamma(2n+1)} = \frac{\Gamma(1/2)\Gamma(n+1)}{4^n\Gamma(n+1/2)} = \frac{(1)_n}{4^n(1/2)_n} $$ by the duplication formula, where $(a)_n = a(a+1)\dotsm(a+n-1)$. Then $$ \sum_{n=0}^{\infty} \binom{2n}{n}^{-k} = \sum_{n=0}^{\infty} \frac{((1)_n)^k}{((1/2)_n)^k} (4^{-k})^n = \pi^{k} {}_{k+1}F_k \left(\begin{matrix} 1,1,1,\dotsc,1 \\ 1/2,1/2,\dotsc,1/2 \end{matrix} ; \frac{1}{4^k} \right), $$ by the definition $$ {}_pF_q \left(\begin{matrix} a_1,a_2,\dotsc,a_p \\ b_1,b_2,\dotsc,b_q \end{matrix} ; z \right) = \sum_{k=0}^{\infty} \frac{(a_1)_n(a_2)_n \dotsm (a_p)_n}{(b_1)_n(b_2)_n \dotsm (b_q)_n} \frac{z^n}{n!}. $$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Now use this for $k=2$. $\endgroup$ – marty cohen Apr 19 '18 at 1:11
  • $\begingroup$ @martycohen Indeed! The OP did ask for a general formula, possibly involving hypergeometric functions, not a useful formula... $\endgroup$ – Chappers Apr 19 '18 at 1:14
  • 1
    $\begingroup$ That's why I upvoted your answer. $\endgroup$ – marty cohen Apr 19 '18 at 2:02
  • $\begingroup$ It'd certainly be nicer if $$\sum_n x^{\binom{2n}{n}}$$ had a closed form we could just integrate a few times. $\endgroup$ – Chappers Apr 19 '18 at 2:06
  • $\begingroup$ That's a lacunary series, since the exponents grow faster than linearly, so almost certainly not. en.wikipedia.org/wiki/Lacunary_function $\endgroup$ – marty cohen Apr 19 '18 at 2:39
3
$\begingroup$

Chappers already wrote these series as hypergeometric functions, so I will just outline how to deal with the case $k=2$, exhibiting a relation with elliptic integrals. The key identities are $$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx,\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2 \binom{2n}{n}} $$ which have been deeply exploited also here.

The case $k=2$. $$ \sum_{n\geq 0}\binom{2n}{n}^{-2}=\phantom{}_2 F_1\left(1,1,1;\tfrac{1}{2},\tfrac{1}{2};\tfrac{1}{16}\right) $$ $$ 2\arcsin^2\left(\frac{x}{4}\right)=\sum_{n\geq 1}\frac{x^{2n}}{4^n n^2 \binom{2n}{n}},\quad \frac{x\arcsin(x/4)}{\sqrt{1-x^2/16}}=\sum_{n\geq 1}\frac{2 x^{2n}}{4^n n \binom{2n}{n}}$$ $$\frac{x^3}{4-\frac{x^2}{4}}+\frac{x^4\arcsin(x/4)}{16\left(1-x^2/16\right)^{3/2}}+\frac{x^2\arcsin(x/4)}{\sqrt{1-x^2/16}}= \sum_{n\geq 1}\frac{4 x^{2n+1}}{4^n \binom{2n}{n}} $$

$$ \frac{x^2 \left(x \sqrt{16-x^2} \left(64-x^2\right)+16 \left(32+x^2\right) \arcsin\left(\frac{x}{4}\right)\right)}{\left(16-x^2\right)^{5/2}}= \sum_{n\geq 1}\frac{(2n+1) x^{2n+1}}{4^n \binom{2n}{n}} $$

$$ \sum_{n\geq 1}\binom{2n}{n}^{-2}=\\=\int_{0}^{\pi/2}\frac{\sin^2(\theta) \left(\sin(\theta) \sqrt{16-\sin^2(\theta)} \left(64-\sin^2(\theta)\right)+16 \left(32+\sin^2(\theta)\right) \arcsin\left(\frac{\sin\theta}{4}\right)\right)}{\left(16-\sin^2\theta\right)^{5/2}}\,d\theta\\ = \frac{3}{5}-\frac{32}{5\sqrt{15}}\arcsin\left(\frac{1}{4}\right)+ 16\int_{0}^{1/4}\frac{u^2 \left(2+u^2\right) \arcsin\left(u\right)}{\sqrt{1-16 u^2}\left(1-u^2\right)^{5/2}}\,du $$ where the last integral is related to $E\left(\frac{1}{4}\right)$ and $K\left(\frac{1}{4}\right)$, which can be computed through algorithms with quadratic convergence (Brent-Salamin and the like).

| cite | improve this answer | |
$\endgroup$

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.