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How would one go about computing the following?:

$$\sum_{n=0}^\infty (.5)^{2n+1} \cdot \frac{{2n}\choose{n}}{n+1}$$

The motivation is that this gives the probability that a random walk on a number line will hit the point $0$ given that we move with probability $.5$ left or right starting at $1$. Wolfram tells me it's $1$, how does one prove it?

In fact, let's go further while we're at it! How can one compute the following?:

$$\sum_{n=0}^\infty (.5)^{2n+1} \cdot \frac{{2n}\choose{n}}{n+1} \cdot (2n+1)$$

This gives us the expected time the aformentioned random walk would take (at the rate of one second per move).

EDIT: Hmm... by the comparison test, it seems like the second summation diverges... therefore, the walk would take an infinite amount of time?

AER
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Raj
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2 Answers2

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The binomial theorem gives the expansion $$ (1-x)^{-1/2}=\sum_{n=0}^\infty\binom{2n}{n}\frac{x^n}{4^n}\tag{1} $$ integrating $(1)$ and dividing by $2$ gives $$ 1-(1-x)^{1/2}=\sum_{n=0}^\infty\binom{2n}{n}\frac{x^{n+1}}{(n+1)2^{2n+1}}\tag{2} $$ Plugging $x=1$ into $(2)$ yields $$ \sum_{n=0}^\infty\binom{2n}{n}\frac1{(n+1)2^{2n+1}}=1\tag{3} $$ which verifies the probability distribution.

$$ \begin{align} \sum_{n=0}^\infty\binom{2n}{n}\frac{2n+1}{(n+1)2^{2n+1}} &=\sum_{n=0}^\infty\binom{2n}{n}\frac{2n+2-1}{(n+1)2^{2n+1}}\\ &=\sum_{n=0}^\infty\binom{2n}{n}\frac1{4^n}-\sum_{n=0}^\infty\binom{2n}{n}\frac1{(n+1)2^{2n+1}}\\ &=(1-1)^{-1/2}-\left(1-(1-1)^{1/2}\right)\\[9pt] &=\infty\tag{4} \end{align} $$ Thus, the second series diverges.

More About Divergence

Asymptotically, $$ \binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{5} $$ Therefore, asymptotically $$ \binom{2n}{n}\frac{2n+1}{(n+1)2^{2n+1}}\sim\frac1{\sqrt{\pi n}}\tag{6} $$ Thus, the second sum diverges.

robjohn
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The generating function for Catalan numbers is:

$$\frac{2}{1+\sqrt{1-4x}}=\sum_{i=0}^\infty C_ix^n$$

Can you do the rest? Hint, plug in $x^2$ first. To derive the generating function, show that

$$C_{n+1}=\sum_{i=1}^nC_kC_{n-k}$$

The second sum diverges.

Alex R.
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  • Unfortunately, my knowledge of generating functions is extremely limited-- i.e., I only know them by name. Can you explain how you obtain at a generating function of a sequence as well as how I can proceed? Thank you so much. – Raj Oct 24 '13 at 03:16
  • See here: http://mikespivey.wordpress.com/2013/03/19/the-catalan-numbers-from-their-generating-function/ – Alex R. Oct 24 '13 at 03:21
  • @AlexR.: Asymptotically, $\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$. Therefore, the terms of the second series are like $\frac1{\sqrt{\pi n}}$. Thus, the second series diverges. – robjohn Oct 24 '13 at 15:19
  • @robjohn: Whoops. Thanks for the correction! – Alex R. Oct 24 '13 at 16:57