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In exercise 36 Miscellaneous Taylor Coefficients using Bernoulli numbers on pages 88-89 of Louis Comtet's Advanced Combinatorics, 1974, one is asked to obtain the following explicit formula for the Bernoulli numbers:

$$B_{2n}=(-1)^{n-1}\dfrac{1+\left[ \varphi _{n}\right] }{2(2^{2n}-1)},$$


$$\varphi _{n}=\dfrac{2(2^{2n}-1)(2n)!}{2^{2n-1}\pi ^{2n}}\displaystyle\sum_{k=1}^{3n}\dfrac{1}{k^{2n}}$$

(with $\displaystyle\sum_{n\geq 0}B_{n}\dfrac{t^{n}}{n!}=\dfrac{t}{e^{t}-1}$), and to prove, among other sums, that

$$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{\dbinom{2n}{n}}=\dfrac{1}{3}+\dfrac{2\pi\sqrt{3}}{27}.\qquad (\ast )$$

Alfred van der Poorten wrote here (section 10): seeing that

$$\displaystyle\sum_{n=1}^{\infty}\dfrac{x^{2n}}{n^{2}\dbinom{2n}{n}}=2\arcsin^{2}\left( \dfrac{x}{2}\right) \qquad (\ast \ast )$$

(...) formula [ $(\ast )$ ] become[s] quite accessible to proof."

I am not able to show formula $(\ast \ast )$ neither how it can be used to prove $(\ast )$.

Question: Could you provide (a) more detailed hint(s) on how and/or different ways in which formula $(\ast )$ can be derived?

Added: For information the other sums are:

$$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n\dbinom{2n}{n}}=\dfrac{\pi \sqrt{3}}{9},\quad\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{2}\dbinom{2n}{n}}=\dfrac{\pi ^{2}}{18},\quad\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{4}\dbinom{2n}{n}}=\dfrac{17\pi ^{4}}{3240}.$$

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you'll find several other related sums over at. – user1709 Nov 14 '10 at 22:54
up vote 17 down vote accepted

Another one, which has been mentioned multiple times on this site (search for Wallis' product) is the following:

$$\displaystyle \int_{0}^{\pi/2} \sin^{2n-1} x \ dx = \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} \ \ n \gt 1$$

Now $$\displaystyle \dfrac{2 \cdot 4 \cdot 6 \cdots (2n-2)}{1 \cdot 3 \cdot 5 \cdots (2n-1)} = \dfrac{2^{2n}}{2n{2n \choose n}} = \dfrac{2^{2n-1}}{n{2n \choose n}}$$


$$\displaystyle \int_{0}^{\pi/2} n\left(\dfrac{\sin x}{2}\right)^{2n-1} \ dx = \dfrac{1}{{2n \choose n}}$$

Now the sum can be found easily.

The formula by Alfred Van der Pooten can be proved using this approach too.

A proof of Wallis Product is given here:

For a (closely related) problem on this site using this, see here: Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$

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+1 for the double hint. – Américo Tavares Nov 12 '10 at 23:49

$1/\binom{2n}{n} = (2n+1) \int_0^1 x^n(1-x)^n dx$. Sum these over $n$ to get the integral of a rational function.

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+1 for the hint. – Américo Tavares Nov 12 '10 at 21:29
BTW, where to find the proof of this formula for integral of $x^n(1-x)^n$? – TCL Nov 12 '10 at 22:07
@TCL: I give a probabilistic proof here: – Qiaochu Yuan Nov 12 '10 at 22:18
@TCL: it is a value of the Beta function (integral). – T.. Nov 13 '10 at 16:01

I'm surprised no one has yet posted the generating function approach. An excellent reference is Sprugnoli, "Sums of Reciprocals of the Central Binomial Coefficients," Integers, Article A27, 2006.

He proves that the generating function of $4^n \binom{2n}{n}^{-1}$ is

$$Z(t) = \frac{1}{1-t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}} + \frac{1}{1-t}.$$ Substituting $t=1/4$ immediately yields $$\sum_{n=0}^{\infty} \binom{2n}{n}^{-1} = \dfrac{4}{3}+\dfrac{2\pi\sqrt{3}}{27}.$$

He also derives generating functions for $\frac{4^n}{n} \binom{2n}{n}^{-1}$, $\frac{4^n}{n^2} \binom{2n}{n}^{-1}$, and several other similar expressions involving the reciprocals of the central binomial coefficients, which allows him to deduce a few dozen expressions for various finite and infinite sums involving $\binom{2n}{n}^{-1}$. Again, it's an excellent reference.

Incidentally, in the introduction the author states that his motivation for writing the paper was the very exercise in Comtet that motivated this question.

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You want to get rid of the factor of $n^2$ in the denominator. So differentiate with respect to $x$, that gets rid of one factor. If you modify what you get you can differentiate again to get rid of the other.

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+1 for the additional help. – Américo Tavares Nov 12 '10 at 22:08

Just as additional information, the inverse of these binomial coefficients satisfy several wonderful identities. For example, on Mathworld I saw the following amazing identity (valid for $k\ge 1$):

$$\sum_{n \ge 1}\frac{1}{n^k \binom{2n}{n}} = \frac{1}{2}{}_{k+1}F_k(\underbrace{1,\ldots,1}_{k+1}; \frac{3}{2}, \underbrace{2,\ldots,2}_{k-1}; \frac{1}{4}),$$

where ${}_{k+1}F_k$ is a generalized hypergeometric function.

How to derive the above result? I don't recall the answer immediately, but perhaps browsing Zeilberger's website might be helpful.

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+1 for pointing out the relation to the generalized hypergeometric function. – Américo Tavares Nov 14 '10 at 10:39
Unfortunately, as stated in the page you linked to, the hypergeometric formula is useful only for $k > 0$ ; remember also that ${}_1 F_0 (a;;z)$ takes only two arguments, and thus the formula in your answer certainly cannot apply. – J. M. Nov 14 '10 at 12:48
thanks for pointing out that $k\ge 1$ is needed; fixed. – user1709 Nov 14 '10 at 22:29
So the relation is applicable only to the other sums I added. – Américo Tavares Nov 14 '10 at 22:36

Here is some more useful info related to your first sum.

Recall that when summing $\sum_{n\ge 0} t_n$ we can use the idea of hypergeometrics, by computing the ratio $t_{n+1}/t_n$, and then reading-off from it the Hypergeometric function that solves the summation.

In your case, the ratio is $$\frac{t_{n+1}}{t_n} = \frac{(n+1)(n+1)}{(n+1/2)4},$$ so the sum over all $t_n$ ranging from $n\ge 0$ instead of $n\ge 1$ as you ask, is nothing but the hypergeometric function $${}_2F_1(1, 1; \frac{1}{2}; \frac{1}{4})$$

which may be verified using some of the ideas above to be equal to $$\frac{2}{27} \left(18+\sqrt{3} \pi \right)$$

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How does this help? – Bonanza Aug 28 '11 at 22:17

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