I was trying to calculate a probability that includes an infinite possibility of events, and the overall probability is expressed by the following formula :

$$P(p) = \sum_{n=0}^{\infty} \frac{(2n)!}{n!(n+1)!} \cdot p^n (1-p)^n = \sum_{n=0}^{\infty} P_n $$ where $p$ is a probability of an specific event and $p \in ]0,1[$.

I used the ratio test and checked that the sum converges for any $p$ of my interest (except for $p=0.5$, where the test is inconclusive). I then tried to get a simpler formula for $P$ that does not involve a sum, as is sometimes possible.

I tried to use the derivative, so I could get something like an ODE, like one can do the sum $e^x$. So

$$P' = \sum_{n=1}^{\infty} \frac{(2n)!}{n!(n+1)!} \cdot n \cdot(1-2p)\cdot p^{n-1}(1-p)^{n-1} = \frac{1-2p}{p(1-p)} \cdot \sum_{n=0}^{\infty} nP_n$$

I tried to express it in function of $P_{n-1}$.

$$P' = (1-2p) \sum_{n=1}^{\infty} \frac{(2n)(2n-1)}{(n+1)n} \cdot \frac{(2(n-1))!}{(n-1)!n!} \cdot n \cdot p^{n-1}(1-p)^{n-1} $$

$$P' = (1-2p) \sum_{n=1}^{\infty} \frac{(4n^2-2n)}{(n+1)} \cdot P_{n-1}$$ $$P' = (1-2p) \sum_{n=0}^{\infty} \frac{(4n^2+6n+2)}{(n+2)} \cdot P_{n}$$

Here I can do a polynomial division, it gives me $4n-2 + 6/(n+2)$

$$P' = (1-2p) \sum_{n=0}^{\infty} \left(4n -2 +\frac{6}{n+2}\right) \cdot P_{n}$$

$$P' = (1-2p) \left[4 \left(\sum_{n=0}^{\infty} nP_n\right) - 2 P + 6 \left( \sum_{n=0}^{\infty} \frac{P_n}{n+2}\right)\right]$$

$$P' = (1-2p) \left[4 \cdot \frac{p(1-p)}{1-2p} \cdot P' - 2 P + 6 \left( \sum_{n=0}^{\infty} \frac{P_n}{n+2}\right)\right]$$

$$[1 - 4p(1-p)]P' = -(2-4p)P + (6-12p) \left( \sum_{n=0}^{\infty} \frac{P_n}{n+2}\right)$$

The expression is already a bit complicated and I don't know how to work with $P_n/(n+2)$. In addition I verified Wolfram Alpha and it gives the following simplification, but with no explanation:

$$P = \frac{|2p-1| - 1}{2(p-1)}$$

Do you suggest something I can do to obtain the previous expression ? Is my approach any good and should I try to go further ? If so how can I work with $P_n/(n+2) ?$

And finally, is there any other way to get the general expression of a power series other than using the derivative?

Thank you.


2 Answers 2


Let $x:=p(1-p)$. We have $$P(p):=\sum_{n=0}^\infty\frac{(2n)!}{n!(n+1)!}\,p^n(1-p)^n=\sum_{n=0}^\infty \binom{2n}n\frac{x^n}{n+1}.$$ Note the standard power series $$\sum_{n=0}^\infty \binom{2n}nx^n=\frac1{\sqrt{1-4x}}$$ which is convergent for $|x|<\frac14$, with antiderivative $$\sum_{n=0}^\infty\binom{2n}n\frac{x^{n+1}}{n+1}=\frac12\Bigl(1-\sqrt{1-4x}\Bigr).$$ Therefore, if $p\neq\frac12$ (so that $x<\frac14$), $$P(p)=\frac1{2p(1-p)}\Bigl(1-\sqrt{1-4p(1-p)}\Bigr)=\frac{1-\sqrt{(1-2p)^2}}{2p(1-p)}=\frac{1-|1-2p|}{2p(1-p)}.$$ This identity is still valid for $p=\frac12$ ($x=\frac14$) by Abel's radial convergence theorem (or more simply, monotone convergence).


The power series of $$\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{2n}{n}x^{n}$$

So $$\frac{1}{p(1-p)}\int_{0}^{p(1-p)}\frac{1}{\sqrt{1-4x}}dx=\sum_{n=0}^{\infty}\frac{(2n!)p^{n}(1-p)^{n}}{n!(n+1)!}$$


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