A092053: Value of Continued Fraction [1;1/2,1/3,1/4,...,1/n,...]

[Joseph Biberstine]

A trivial equivalence to clean up the right side:

[1; 1, 1/2, 1/3, 1/4, 1/5, ..., 1/n, ...] = Pi/2

- JRB

Paul D. Hanna wrote:
> Seqfans, 
>       My guess is: 
>  
> (*) [1;1/2,1/3,1/4,...,1/n,...] = 1/(Pi/2 - 1) = 1.7519383938841...
>  
> which would explain the behavior of the numerators and 
> denominators of the convergents. 
>  
> Proof of (*) anyone? 
>  
> On Wed, 9 Aug 2006 19:53:56 -0400 "Paul D. Hanna" <pauldhanna at juno.com>
> writes:
>> Seqfans, 
>> Can anyone evaluate the continued fraction:
>>    x = [1;1/2,1/3,1/4,...,1/n,...].
>>  
>> Convergence is very slow: 
>> at 500000 partial quotients, x = 1.751943215111853159301... ;
>> at 600000 partial quotients, x = 1.751942411573727042118...
>>  
>> The convergents of the CF are interesting, and are related to Pi/2.
>>  
>> See A092053: 
>> "Denominators of the convergents of the continued fraction 
>> expansion
>> [1;1/2,1/3,1/4,...,1/n,...]."
>> COMMENT: 
>> "Numerators of convergents are A001902 (successive denominators 
>> of Wallis's product approximation to Pi/2). 
>> Sum of numerators and denominators equals powers of 2: 
>> A001902(n) + a(n) = 2^A092054(n)."
>>  
>> Thanks, 
>>    Paul