# How to compute this series: $\sum_{k=0}^\infty \frac{C_k}{2^{2k+1}}$ [closed]

How to compute this series: $$\sum_{k=0}^\infty \frac{C_k}{2^{2k+1}}$$ where $$C_k$$ is the catalan number: $$C_k=\frac{1}{k+1}{2k \choose k}$$. (Further, is there any general method to treat this question with general $$C_k$$?)

## closed as off-topic by abx, j.c., მამუკა ჯიბლაძე, Boris Bukh, Max AlekseyevJan 1 at 17:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – abx, j.c., მამუკა ჯიბლაძე, Boris Bukh, Max Alekseyev
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• Use the power series for $\arcsin$. Not appropriate for MO. – abx Dec 31 '18 at 18:21
• The Taylor series for $\arcsin$ together with the binomial formula for $C_n$ does the trick, however, recognizing $\arcsin$ is not trivial at all. I therefore vote to reopen. – Jan-Christoph Schlage-Puchta Jan 2 at 9:08

## 4 Answers

The ordinary generating function of the $$C_k$$ is well-known:

$$\sum_{k \geq 0} C_k x^k = \frac{1 - \sqrt{1 - 4x}}{2x}$$

This can be deduced from the problem of counting binary planar trees and using familiar generatingfunctionology techniques. From there the problem is not difficult.

$$k$$-th summand is the probability that a symmetric random walk with steps $$\pm 1$$ returns first time to the initial point at time $$2(k+1)$$. Since this random walk is recurrent, the sum of probabilities equals 1.

• @mathworker21 There are many independent proofs of recurrence, which work in more general setting. – Fedor Petrov Jan 1 at 8:02

And as it often happens when the sum of a series is rational, it is simply telescopic: $$\frac{C_k}{2^{2k+1}}=\frac{\binom{2k}k}{4^k}-\frac{\binom{2 (k+1) }{k+1} }{4^{k+1}}.$$

Maple 2018.2.1 and Mathematica 11.3.0 are your friends

sum(binomial(2*k, k)/((k+1)*2^(2*k+1)), k = 0 .. infinity);


$$1$$

Sum[Binomial[2*k, k]/(k + 1)/(2^(2*k + 1)), {k, 0, Infinity}]


$$1$$

sum(binomial(2*k, k)*x^k/(k+1), k = 0 .. infinity, parametric);


$$\cases{2\, \left( 1+\sqrt {1-4\,x} \right) ^{-1}&4\, \left| x \right| \leq 1\cr \infty &1<4\,x\cr \sum _{k=0}^{\infty }{\frac {{2\,k\choose k}{x}^{k}}{k+1}}&otherwise\cr}$$

Sum[Binomial[2*k, k]/(k + 1)*x^k, {k, 0, Infinity},GenerateConditions -> True]


$$\text{ConditionalExpression}\left[\frac{1-\sqrt{1-4 x}}{2 x},x=-\frac{1}{4}\lor x=\frac{1}{4}\lor \left| x\right| <\frac{1}{4}\right]$$

Edit. MMA code is added.