Sum of reciprocals of double factorials

Jonathan Post jvospost3 at gmail.com
Mon Aug 4 21:41:16 CEST 2008

Very elegant, Max Alekseyev!

The generalization to k-tuple factorials is similar, and the
convergents require a modulo k adjustment?

In my earlier email, of course I meant "What are the real numbers to
which the sum of reciprocals of k-tuple
factorials converges?"

With your formulae, is this now worth submitting, as definitive rather
than merely an analogue of existing seqs?

Best,

Jonathan Vos Post

On Mon, Aug 4, 2008 at 12:35 PM, Max Alekseyev <maxale at gmail.com> wrote:
> Corrections:
>
> SUM[i=0..oo] 1 / i!! = (sqrt(Pi/2) * erf(1/sqrt(2))+1) * sqrt(e) ~= 3.0594074
>
> and
>
> to get an expression for a(n) we need to consider the terms on odd and
> even positions separately:
>
> SUM[i=0..oo] x^(2i) / (2i)!! = exp(x^2/2);
>
> SUM[i=0..oo] x^(2i+1) / (2i+1)!! = sqrt(Pi/2) * erf(x/sqrt(2)) * exp(x^2/2).
>
> Then
>
> a(2n) = floor( sqrt(e) * (2n)!! ) / (2n)!! + floor( sqrt(e*Pi/2) *
> erf(1/sqrt(2)) * (2n-1)!! ) / (2n-1)!!
> ~= floor( 1.648721271 * (2n)!!) / (2n)!! + floor( 1.410686134 *
> (2n-1)!! ) / (2n-1)!!
>
> and
>
> a(2n+1) = floor( sqrt(e) * (2n)!! ) / (2n)!! + floor( sqrt(e*Pi/2) *
> erf(1/sqrt(2)) * (2n+1)!! ) / (2n+1)!!
> ~= floor( 1.648721271 * (2n)!!) / (2n)!! + floor( 1.410686134 *
> (2n+1)!! ) / (2n+1)!!
>
> Regards,
> Max
>
> On Mon, Aug 4, 2008 at 12:26 PM, Max Alekseyev <maxale at gmail.com> wrote:
>> Note that
>>
>> SUM[i=0..oo] x^i / i!! = (sqrt(Pi/2) * erf(x/sqrt(2))+1) * exp(x^2/2)
>>
>> so that
>>
>> SUM[i=0..oo] x^i / i!! = (sqrt(Pi/2) * erf(1/sqrt(2))+1) * sqrt(e) ~= 3.0594074
>>
>> and your sequence can be defined as
>>
>> a(n) = floor( (sqrt(Pi/2) * erf(1/sqrt(2))+1) * sqrt(e) * n!! ) / n!!
>> ~= floor( 3.0594074 * n!! ) / n!!
>>
>> Regards,
>> Max
>>
>> On Mon, Aug 4, 2008 at 11:55 AM, Jonathan Post <jvospost3 at gmail.com> wrote:
>>> Is this worth adding to OEIS?
>>>
>>> Sum of reciprocals of double factorials
>>> (2 seqs "frac" for numerators and denominators)
>>>
>>> SUM[i=0..n] 1/A006882(i)
>>>
>>> n   numerator/denominator
>>> 0   1/0!! = 1/1
>>> 1   1/0!! + 1/1!! = 2/1
>>> 2   1/0!! + 1/1!! + 1/2!! = 5/2
>>> 3   1/0!! + 1/1!! + 1/2!! + 1/3!! = 17/6
>>> 4   1/0!! + 1/1!! + 1/2!! + 1/3!! + 1/4!! = 71/24
>>> 5   1/0!! + 1/1!! + 1/2!! + 1/3!! + 1/4!! + 1/5!! = 121/40
>>> 6   1/0!! + 1/1!! + 1/2!! + 1/3!! + 1/4!! + 1/5!! + 1/6!! = 731/240
>>> 7   1/0!! + 1/1!! + 1/2!! + 1/3!! + 1/4!! + 1/5!! + 1/6!! + 1/7!! =
>>> 1711/560 ~ 3.0553571
>>> 8   1/0!! + 1/1!! + 1/2!! + 1/3!! + 1/4!! + 1/5!! + 1/6!! + 1/7!! +
>>> 1/8!! = 41099/13440 ~ 3.05796131
>>>
>>> The series obviously converges (being of order 1/n^2).
>>>
>>> This is to double factorials A006882 as A007676/A007677is to factorial.
>>>
>>> The WIMS continued fraction online calculator seems to be unavailable
>>> at the moment, O i've stopped with the above by-hand draft.
>>>
>>> If this is of interest, then there would be an array A[k,n] = nth
>>> convergent to sum of reciprocals of the k-th multiple factorial, using
>>> the correct definitions by njas, Robert G. Wilson v, Mira Bernstein of
>>> k-th multiple factorial.
>>>
>>> What is the real number to which the sum of reciprocals of double
>>> factorials converges?
>>>
>>> What are the real numbers to which the sum of reciprocals of double
>>> factorials converges?
>>>
>>
>

By my calculation, A003242(n) should tend to k^n, where k is given by:

By numerical methods I find k ~= 1.750241291718309, can anyone suggest
how to solve that equation to get a closed form?

Actual values seem to support this; here are some selected cases:

====================
16384: 1.7501574922

Hugo