Sum of reciprocals of double factorials

Max Alekseyev maxale at gmail.com
Mon Aug 4 21:35:52 CEST 2008

```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?
>>
>

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