How to prove: $$1+\frac13 \left(1+\frac{1}{5}\left(1+\frac{1}{7}\left(1+\frac{1}{9}\left(1+\dots \right) \right) \right) \right)=\sqrt{\frac{\pi e}{2}} \text{erf} \left( \frac{1}{\sqrt{2}} \right)$$

Since we can represent the series for $e$ in a pretty form:

$$1+\frac12 \left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\dots \right) \right) \right) \right)=e-1$$

I've been trying to search closed forms for some similar expressions. For example, it's easy to see that:

$$1+\frac12 \left(1+\frac{1}{4}\left(1+\frac{1}{6}\left(1+\frac{1}{8}\left(1+\dots \right) \right) \right) \right)=\sqrt{e}$$

However, I don't know how to prove the expression in the title. Seems to be something for Ramanujan to consider.

P.S. the closed form was given by Wolfram Alpha.

See a general formula provided by Lucian in the comments.

WA also provides another general formula:

$$\sum_{n=0}^\infty \frac{1}{\prod_{k=0}^n bk+c}=e^{\frac{1}{b}} \left(\frac{1}{b} \right)^{1-\frac{c}{b}} \left( \Gamma\left(\frac{c}{b} \right) -\Gamma\left(\frac{c}{b},\frac{1}{b} \right) \right)$$

Still don't know how to prove it, but I've noticed the emergence of the hypergeometric functions in some cases, which makes it possible to transform the expression to hypergeometric series.

Yuriy S
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    You can look here http://math.stackexchange.com/questions/833920/an-infinite-series-plus-a-continued-fraction-by-ramanujan – E. Joseph Jan 09 '17 at 12:54
  • In general, $$\sum_{n=0}^\infty~2^n\cdot\frac{x^{2n+1}}{(2n+1)!!} ~=~ \frac{\sqrt\pi}2\cdot e^{x^2}\cdot\text{erf}(x).$$ – Lucian Jan 09 '17 at 12:58
  • Start by writing the well-known [Taylor series](http://en.wikipedia.org/wiki/Taylor_series#List_of_Maclaurin_series_of_some_common_functions) expansion for [exponential function](http://en.wikipedia.org/wiki/Characterizations_of_the_exponential_function#Characterizations), then integrate it term by term to obtain that of the [error function](http://en.wikipedia.org/wiki/Error_function), and lastly evaluate their [Cauchy product](http://en.wikipedia.org/wiki/Cauchy_product). – Lucian Jan 09 '17 at 14:10
  • @Lucian, thanks for the advice, but this looks complicated. Maybe using hypergeometric functions will be easier for me – Yuriy S Jan 09 '17 at 14:12
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    The link from @E.Joseph has your answer. – Paramanand Singh Jan 09 '17 at 14:21
  • @ParamanandSingh, I see now, yes. However, for some reason it's the continued fraction which gives this value, not the series, according to Nikos Bagis's answer. I'll consider everything more carefully to see what I missed – Yuriy S Jan 09 '17 at 14:25
  • The LHS of your question is equal to $f(1)$ of the linked question whose value is $$\sqrt{e} \int_{0}^{1}e^{-t^{2}/2}\,dt=\sqrt{2e}\int_{0}^{1/\sqrt{2}}e^{-t^{2}}\,dt=\sqrt{\frac{\pi e} {2}}\operatorname {erf}\left(\frac{1}{\sqrt{2}}\right)$$ – Paramanand Singh Jan 09 '17 at 15:46
  • BTW expansions as in your post are called Engel expansions and each positive real number has a unique expansion of this kind. See https://en.wikipedia.org/wiki/Engel_expansion – Paramanand Singh Jan 09 '17 at 15:52

1 Answers1


Using Euler's Beta function,

$$ S = \sum_{n\geq 0}\frac{1}{(2n+1)!!} = \sum_{n\geq 0}\frac{2^n n!}{(2n+1)!} = \sum_{n\geq 0} \frac{2^n}{n!} B(n+1,n+1) \tag{1}$$ leads to: $$ S = \int_{0}^{1}\sum_{n\geq 0}\frac{\left(2x(1-x)\right)^n}{n!}\,dx = \int_{0}^{1}\exp\left(2x(1-x)\right)\,dx\tag{2}$$ and the last integral is easy to convert into the $\text{Erf}$ format through the substitution $x=\frac{1}{2}+t$.

Jack D'Aurizio
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