Evaluating $\sum_{n=i}^{\infty} {2n \choose n-i}^{-1}$

Question

$\displaystyle \sum_{n=i}^{\infty} {2n \choose n-i}^{-1}=\sum_{n=i}^{\infty} \frac {1}{{2n \choose n-i}}$ is a very interesting one. Here is what I have from WolframAlpha.

$$\displaystyle \sum_{n=0}^{\infty} {2n \choose n}^{-1}=\frac{2}{27}(18+\sqrt{3}\pi)$$

$$\displaystyle \sum_{n=1}^{\infty} {2n \choose n}^{-1}=\frac{1}{27}(9+2\sqrt{3}\pi)$$

$$\displaystyle \sum_{n=1}^{\infty} {2n \choose n-1}^{-1}=\frac{1}{27}(9+5\sqrt{3}\pi)$$

For $i \geq 2$, WA just comes up with closed forms involving generalised hypergeometric functions. Here is one example. I would conjecture that this is still in fact expressible in terms of the sort of form we see above but I have no clue about hypergeometric functions so I was hoping somebody could enlighten me. Also, it looks like some sort of computational artefact of not being able to start the sum at $n=1$. It seems to give a natural, consistent form for the first two cases ($i=0,1$), and when asking WA to sum it in the case $i=1$, but starting from $n=2$, it uses a hypergeometric function in the answer, whereas we know the answer should in fact be $\frac{1}{27}(-18+5\sqrt{3}\pi)$.

Also, could anybody come up with a (more elementary i.e. anything not involving special functions) proof of the formula $\displaystyle \sum_{n=1}^{\infty} {2n \choose n}^{-1}=\frac{1}{27}(9+2\sqrt{3}\pi)$?

I have struggled and failed, and this is the part I would most like answered if possible.

Thanks, and good luck (if needed).

Answer 1

Generating Function for the Reciprocals of the Central Binomial Coefficients

Using the fact that $$ \frac{4^n}{\binom{2n}{n}} =\frac{n}{n-\frac12}\frac{4^{n-1}}{\binom{2n-2}{n-1}}\tag{1} $$ we can compute the generating function for the reciprocals of the central binomial coefficients: $$ \begin{align} \frac{4^n}{\binom{2n}{n}} &=\prod_{k=1}^n\frac{k}{k-\frac12}\\ &=\frac{\Gamma(n+1)}{\Gamma(1)}\frac{\Gamma(\frac12)}{\Gamma(n+\frac12)}\\ &=n\,\mathrm{B}(n,\tfrac12)\\ &=n\int_0^1(1-t)^{n-1}t^{-1/2}\,\mathrm{d}t\\ &=2n\int_0^1(1-t^2)^{n-1}\,\mathrm{d}t\\ \frac{x^n}{\binom{2n}{n}} &=\frac{nx}2\int_0^1\left(\frac{x(1-t^2)}4\right)^{n-1}\,\mathrm{d}t\\ \sum_{n=1}^\infty\frac{x^n}{\binom{2n}{n}} &=\frac{x}2\int_0^1\frac1{\left(1-\frac{x(1-t^2)}4\right)^2}\,\mathrm{d}t\\ &=\frac{8x}{(4-x)^2}\int_0^1\frac1{\left(1+\frac{x}{4-x}t^2\right)^2}\,\mathrm{d}t\\ &=\frac{8x^{1/2}}{(4-x)^{3/2}}\int_0^{\sqrt{\frac{x}{4-x}}}\frac1{\left(1+t^2\right)^2}\,\mathrm{d}t\\ &=\frac{4x^{1/2}}{(4-x)^{3/2}}\left[\frac{t}{1+t^2}+\tan^{-1}(t)\right]_0^{\sqrt{\frac{x}{4-x}}}\\ &=\frac{x}{4-x}+\frac4{4-x}\sqrt{\frac{x}{4-x}}\tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right)\\ \sum_{n=0}^\infty\frac{x^n}{\binom{2n}{n}} &=\frac4{4-x}\left[1+\sqrt{\frac{x}{4-x}}\tan^{-1}\left(\sqrt{\frac{x}{4-x}}\right)\right]\\ &=\frac4{4-x}\left[1+\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\right]\tag{2} \end{align} $$ Plug in $x=1$ and we get $$ \begin{align} \sum_{n=0}^\infty\frac1{\binom{2n}{n}} &=\frac43\left[1+\sqrt{\frac13}\sin^{-1}\left(\frac12\right)\right]\\ &=\frac43+\frac{2\pi}{9\sqrt3}\tag{3} \end{align} $$

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Extending the Previous Result

Suppose we have $$ F_k(x)=\sum_{n=k}^\infty\frac{x^{n-k}}{\binom{2n}{n-k}}\tag{4} $$ $F_0(x)$ is given in $(2)$ above.

We can use the identity $$ \binom{2n}{n-k-1}=\frac{n-k}{n+k+1}\binom{2n}{n-k}\tag{5} $$ to get $$ \begin{align} \frac1{\binom{2n}{n-k-1}}-\frac1{\binom{2n}{n-k}} &=\frac1{\frac{n-k}{n+k+1}\binom{2n}{n-k}}-\frac1{\binom{2n}{n-k}}\\ &=\frac1{n-k}\left[\frac{n+k+1}{\binom{2n}{n-k}}-\frac{n-k}{\binom{2n}{n-k}}\right]\\ &=\frac{2k+1}{n-k}\frac1{\binom{2n}{n-k}}\tag{6} \end{align} $$ Equation $(6)$ shows that $$ \frac{\mathrm{d}}{\mathrm{d}x}(xF_{k+1}(x)-F_k(x))=\frac{2k+1}{x}(F_k(x)-1)\tag{7} $$ Formula $(7)$ gives us a way to compute $F_{k+1}$ from $F_k$, via integration.

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Example

If we let $x=4\sin^2(\theta)$, then $$ \begin{align} &\int\frac{F_0(x)-1}{x}\mathrm{d}x\\ &=\int\left[\frac1{4-x}+\frac4{4-x}\sqrt{\frac1{x(4-x)}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\right]\mathrm{d}x\\ &=\int\left[\frac1{4\cos^2(\theta)}+\frac1{\cos^2(\theta)}\sqrt{\frac1{16\sin^2(\theta)\cos^2(\theta)}}\,\,\theta\right]8\sin(\theta)\cos(\theta)\,\mathrm{d}\theta\\[3pt] &=2\int\left[\tan(\theta)+\theta\sec^2(\theta)\right]\mathrm{d}\theta\\[12pt] &=2\theta\tan(\theta)-1\\[9pt] &=2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)-1\tag{8} \end{align} $$ where the constant of integration was chosen because $xF_1(x)-F_0(x)=-1$ at $x=0$.

Thus, we get $$ xF_1(x)-F_0(x)=2\sqrt{\frac{x}{4-x}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)-1\tag{9} $$ Therefore, $$ F_1(x)=\frac1{4-x}+\frac{12-2x}{4-x}\frac1{\sqrt{x(4-x)}}\sin^{-1}\left(\frac{\sqrt{x}}2\right)\tag{10} $$ Plug in $x=1$ and we get $$ \sum_{n=1}^\infty\frac1{\binom{2n}{n-1}}=\frac13+\frac{5\pi}{9\sqrt3}\tag{11} $$

Answer 2

Using Maple, I also get $$ \sum _{n=2}^{\infty } {2\,n\choose n-2} ^{-1}={\frac { 23}{6}}-{\frac {13\,\pi \,\sqrt {3}}{27}} $$ $$\sum _{n=3}^{\infty } {2\,n\choose n-3} ^{-1}=\dfrac34+{ \frac {2\,\pi \,\sqrt {3}}{27}} $$ For $i=4$ Maple can't seem to simplify $$\sum _{n=4}^{\infty } {2\,n\choose n-4} ^{-1}= {\mbox{$_3$F$_2$}(1,1,9;\,9/2,5;\,1/4)} $$ but numerically it appears that the answer is $$- \dfrac{211}{60}+\dfrac{23 \pi \sqrt{3}}{27}$$ for $i=5$ it is $$ \dfrac{6169}{840}-\dfrac{31 \pi \sqrt{3}}{27}$$ and for $i=6$ it is $$ {\frac {1709}{2520}}+{\frac {2\,\pi \,\sqrt {3}}{27}}$$ etc. The coefficient of $\sqrt{3} \pi/27$ seems to have the generating function $${\frac {6\,t-6}{ \left( {t}^{2}+t+1 \right) ^{2}}}+{\frac {-5\,t+8}{{t }^{2}+t+1}} $$

EDIT: Ah! Using Jack's method, the generating function for this sequence turns out to be

$$ \eqalign{ \sum_{i=0}^\infty t^i \sum_{n=i}^\infty {2n \choose n-i}^{-1} &= \sum_{i=0}^\infty t^i \sum_{n=i}^\infty \int_0^1 (n+i) t^{n-i} (1-t)^{n+i-1}\; dt\cr =&\dfrac{\ln((1+\sqrt{t})/(1-\sqrt{t})) (t+1) t^{3/2}}{(t^2+t+1)^2} - \dfrac{ \ln(1-t) (t^2 + 3 t + 1) t}{2(t^2+t+1)^2}\cr & - \dfrac{(5 t^3 - 3 t^2 - 9 t - 2) \pi \sqrt{3}}{27 (t^2 + t + 1)^2} + \dfrac{1-t}{3 (t^2 + t + 1)} - \dfrac{1}{t-1} - 1 }$$

Answer 3

$$\sum_{n=0}^{+\infty}\binom{2n}{n}^{-1}=\sum_{n\geq 0}(2n+1)\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}=\sum_{n\geq 0}\int_{0}^{1}(2n+1)(x(1-x))^n\,dx$$ hence: $$\sum_{n=0}^{+\infty}\binom{2n}{n}^{-1}=\int_{0}^{1}\frac{1+x(1-x)}{(1-x(1-x))^2}\,dx$$ and the integral can be easily evaluated through the residue theorem. Other cases are similar.

Answer 4

The basic result is $~\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{\displaystyle{2n\choose n}n^2}=2\arcsin^2x.~$ By repeating the process of differentiation

and multiplication twice, and letting $x=\dfrac12,~$ you will obtain a closed form for your series.

Then, by using similar means, try deducing a general formula for each modified version of

this initial result, for $i>0$, namely, $i=1,2,3$, etc. If WA does not return nice results, use

FullSimplify[...].

Answer 5

We can do an example to see how this calculation actually works. Suppose we seek to evaluate $$\sum_{n\ge 4} {2n\choose n-4}^{-1}.$$

This is $$\sum_{n\ge 4} \frac{(n-4)! \times (n+4)!}{(2n)!} = \sum_{n\ge 4} \frac{\Gamma(n-3) \times \Gamma(n+5)}{\Gamma(2n+1)} \\ = \sum_{n\ge 4} (2n+1) \frac{\Gamma(n-3) \times \Gamma(n+5)}{\Gamma(2n+2)} = \sum_{n\ge 4} (2n+1) \mathrm{B}(n+5, n-3).$$

Recall the beta function integral $$\mathrm{B}(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}} dt.$$

This gives for the sum the representation $$\int_0^\infty \sum_{n\ge 4} (2n+1) \frac{t^{n+4}}{(1+t)^{2n+2}} dt = \int_0^\infty \frac{t^4}{(1+t)^2} \sum_{n\ge 4} (2n+1) \frac{t^{n}}{(1+t)^{2n}} dt \\ = \int_0^\infty \frac{t^4}{(1+t)^2} \times \frac{(9t^2+11t+9)t^4}{(1+t)^6(t^2+t+1)^2} dt \\ = \int_0^\infty \frac{(9t^2+11t+9)t^8}{(1+t)^8(t^2+t+1)^2} dt.$$

This integral can be evaluated by considering $$f(z) = \log z \times \frac{(9z^2+11z+9)z^8}{(1+z)^8(z^2+z+1)^2}$$ evaluated on a keyhole contour with the slot on the positive real axis and the branch cut of the logarithm also on that axis, traversed counterclockwise.

There are four segments: $\Gamma_1$ just above the cut, $\Gamma_2$ the large circle of radius $R$, $\Gamma_3$ the segment below the cut and $\Gamma_4$ the small circle around the origin of radius $\epsilon$.

Using $$|\log(Me^{i\theta})| = |\log M + i\theta| = \sqrt{(\log M)^2+\theta^2}$$ we obtain that the contribution along $\Gamma_2$ is $$2\pi R \times \log R / R^2 \to 0$$ as $R\to \infty,$ so it vanishes.

The contribution along $\Gamma_4$ is $$2\pi\epsilon \times |\log \epsilon| \times \epsilon^8 \to 0$$ as $\epsilon\to 0,$ so it vanishes as well.

We get two contributions from the logarithm below the positive real axis along $\Gamma_3$, one of which cancels the integral along $\Gamma_1$ and the other one of which is $$-2\pi i \int_0^\infty \frac{(9t^2+11t+9)t^8}{(1+t)^8(t^2+t+1)^2} dt$$ i.e. the integral we are trying to calculate.

Let $\rho_0 = -1$ and $$\rho_{1,2} = -\frac{1}{2} \pm \frac{\sqrt{3}i}{2}.$$ We thus have that by the Cauchy Residue Theorem applied to the keyhole contour $$\int_0^\infty \frac{(9t^2+11t+9)t^8}{(1+t)^8(t^2+t+1)^2} dt = - (\mathrm{Res}_{z=\rho_0} f(z) + \mathrm{Res}_{z=\rho_1} f(z) + \mathrm{Res}_{z=\rho_2} f(z)). $$

Now to calculate these residues we use a CAS but it will need some assistance namely from the expansion about $\rho$ $$\log z= \log(\rho + z -\rho) = \log\rho + \log (1 + (z-\rho)/\rho) = \log\rho + \sum_{q\ge 1} \frac{(-1)^{q+1}}{q} \frac{(z-\rho)^q}{\rho^q}.$$

We use this expansion to compute the residues, making sure that the constant term $\log\rho$ agrees with the chosen branch.

This gives for the first residue that $$\mathrm{Res}_{z=\rho_0} f(z) = \frac{57}{20} - 3\pi i$$ and for the second $$\mathrm{Res}_{z=\rho_1} f(z) = \frac{1}{3} + \frac{23\pi \sqrt{3}}{27} + \pi i -\frac{1}{3}\sqrt{3}i$$ and for the third $$\mathrm{Res}_{z=\rho_2} f(z) = \frac{1}{3} - \frac{46\pi \sqrt{3}}{27} + 2\pi i +\frac{1}{3}\sqrt{3}i.$$

Adding these and negating the result we finally have $$\frac{23\pi\sqrt{3}}{27} - \frac{211}{60}.$$