golden mean from an agm.

Let

S2=sum_{n>0} 1/F(2*n-1)

S3=sum_{n>0} 1/L(2*n)

then

agm(S2/sqrt(5),1/4+S3)

=Pi/8/log((1+sqrt(5))/2)

Proof: As before, but now use

the evaluations of the nome

q=exp(-Pi*K(k')/K(k))=(3-sqrt(5))/2

Check:

F(n)=fibonacci(n)

L(n)=if(n,F(2*n)/F(n),2)

S2=suminf(n=1,1./F(2*n-1));

S3=suminf(n=1,1./L(2*n));

print(agm(S2/sqrt(5),1/4+S3)-Pi/8/log((1+sqrt(5))/2));

-2.5860875718090325175134436270000000000 E-37