Mixed-radix maximal digits $\sum_{i=0}^\infty \frac{(2i)i!}{(2i+1)!!} = 2$

Question

In A Spigot Algorithm for the Digits of π by Rabinowitz and Wagon, the following lemma says the "maximal digit" representation $(0;2,4,6,8\dots)$ in the mixed-radix base $(\frac 1 3, \frac 2 5, \frac 3 7, \dots)$ is bounded by 2:

$$0 + \frac 1 3 \left(2 + \frac 2 5 \left( 4 + \frac 3 7 ( 6 + \cdots ) \right) \right) = \sum_{i=0}^\infty \frac{(2i)i!}{(2i+1)!!} = 2$$

They sketch a proof by induction based on remainders: the differences of the partial sum and 2. Is there a direct proof based on combinatorial interpretation or well-known series? The $\pi/2$ formula of $(1;1,1,1,1,\dots)$ is apparently due to Newton and can be derived from the arctan Taylor series.

Answer 1

We have a telescopic series \begin{align*} \sum_{n=0}^{\infty} \frac{n \cdot n!}{(2n+1)!!} &= \sum_{n=0}^{\infty} \frac{(2n+1)n!-(n+1)n!}{(2n+1)!!} \\ &= \sum_{n=0}^{\infty} \Big(\frac{n!}{(2n-1)!!} - \frac{(n+1)!}{(2n+1)!!}\Big)\\ &= \frac{0!}{(-1)!!} - \lim_{n \to +\infty} \frac{(n+1)!}{(2n+1)!}\\ &= 1. \end{align*} We have set $(-1)!! = 1$ to keep the relation $(2n+1)!! = (2n+1)(2n-1)!!$ true when $n=0$.

Answer 2

It will be fascinating to have a combinatorial/probabilistic interpretation of the sum.

Unfortunately, I don't have a good lead in this direction yet. Let me think about this, but in the meantime, here is an analytic solution:

\begin{align*} \sum_{n=0}^{\infty} \frac{n \cdot n!}{(2n+1)!!} &= \sum_{n=0}^{\infty} 2^{n}n \frac{(n!)^2}{(2n+1)!} \\ &= \sum_{n=0}^{\infty} 2^{n}n \int_{0}^{1} x^n(1-x)^n \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{2x(1-x)}{(1 - 2x(1-x))^2} \, \mathrm{d}x \\ &= \left[ \frac{2x - 1}{2(1 - 2x(1-x))} \right]_{0}^{1} = 1. \end{align*}

Now the desired conclusion follows by multiplying both sides by $2$.

Answer 3

Here's a probabilistic interpretation of the equivalent identity $$\sum_{n=0}^\infty \frac{n\cdot n!}{(2n+1)!!},$$ inspired by Christophe Leuridan's telescoping-based answer.

Begin at time $t=0$. At time $t$, pick a uniformly random element of $\{0,1,\ldots,2t\}$. It the element is greater than $t$, terminate; otherwise, continue. It is easy to see that this process terminates with probability $1$. On the other hand, the probability that it terminates exactly at time $t=n$ is $$\left(\prod_{k=0}^{n-1}\Pr\big[\text{does not terminate at $t=k$}\big]\right)\cdot\Pr\big[\text{terminates at $t=n$}\big],$$ which is $$\prod_{k=0}^{n-1}\left(1-\frac k{2k+1}\right)\cdot\frac n{2n+1}=\frac{n\prod_{k=0}^{n-1}(k+1)}{(2n+1)\prod_{k=0}^{n-1}(2k+1)}=\frac{n\cdot n!}{(2n+1)!!}.$$ We conclude that $$\sum_{n=0}^\infty \frac{n\cdot n!}{(2n+1)!!}=\Pr[\text{process terminates}]=1.$$

Answer 4

Another way.

$$\sum_{i=0}^\infty\frac{2i\,i!}{(2i+1)!!}= \sum_{i=0}^\infty\frac{2^{i+1}i(i!)^2}{(2i+1)!}= \sum_{i=0}^\infty \frac{i\,i!}{2^{i-1}(\tfrac32)_{(i)}}\\= 2\sum_{i=0}^\infty \frac{\left[(i+1)!-i!\right]i!}{(\tfrac32)_{(i)}}\tfrac{(\tfrac12)^i}{i!}\\\stackrel{EIF}{=} 2\left(_2F_1(2,1;\tfrac32;\tfrac12)- _2F_1(2,1;\tfrac32;\tfrac12) \right)\\ =\tfrac2{B(1,\tfrac12)}\left(\int_0^1(1-x)^{-1/2}(1-\tfrac12x)^{-2}dx- \int_0^1(1-x)^{-1/2}(1-\tfrac12x)^{-1}dx\right)\\ =\int_0^1\frac{2xdx}{\sqrt{1-x}(2-x)^2}=2. $$ Evaluation of integral is by WolframAlpha.

$EIF:$ Euler Integral Formula for $_2F_1$ hypergeometric function