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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 $C_k$?)

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closed as off-topic by abx, j.c., მამუკა ჯიბლაძე, Boris Bukh, Max Alekseyev Jan 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
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Use the power series for $\arcsin$. Not appropriate for MO. $\endgroup$ – abx Dec 31 '18 at 18:21
  • $\begingroup$ 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. $\endgroup$ – Jan-Christoph Schlage-Puchta Jan 2 at 9:08
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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.

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$k$-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.

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  • 1
    $\begingroup$ @mathworker21 There are many independent proofs of recurrence, which work in more general setting. $\endgroup$ – Fedor Petrov Jan 1 at 8:02
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And as it often happens when the sum of a series is rational, it is simply telescopic: $$ \frac{C_k}{2^{2k+1}}=\frac{\binom{2k}k}{4^k}-\frac{\binom{2 (k+1) }{k+1} }{4^{k+1}}. $$

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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.

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