Surely known, but maybe worthwhile for Eric's mathworld: sum(n=1,infty,1.0/fibonacci(2*n-1)) == sqrt(5)/4*( Th22(be) ) where Th22(q)=4*q * sum( n=0,infty,q^(2*(n^2+n)) )^2 \\ ==Theta2^2(q^2) and be = (sqrt(5)-1)/2; \\ == 0.61803398... sum(n=1,infty,1.0/fibonacci(2*n)) == sqrt(5)*( La22(be) ) where La22(q)=sum(n=1,infty,q^(n*(n+1))/(1-q^(2*n))) \\ Lambert, cf. below and be as before I cannot, however, combine these two into anything neater than just the plain sum (how can one marry Theta with Lambert?). cheers, jj P.S.: some pari/gp code for checking and some additional relations: \\ -------------------------------- default(realprecision,100); default(format,"g.15"); default(echo,1); N=500 \\ sum N terms with linear series S=ceil(sqrt(N)) \\ sum S terms with faster (Theta-type) series Th3(q)=1+2*sum(n=1,S,q^(n^2)) \\ Theta3 Th4(q)=Th3(-q) \\ Theta4 Th2(q)=2*q^(1/4)*sum(n=0,S,q^(n^2+n)) \\ Theta2 \\Th22(q)=(Th3(q)^2-Th4(q)^2)/2 \\ == Theta2^2(q^2) Th22(q)=4*q * sum( n=0,S,q^(2*(n^2+n)) )^2 \\ Theta2^2(q^2) \\La(q)=sum(n=1,N, q^n/(1-q^n)) \\ Lambert series La(q)=sum(n=1,S, q^(n^2)*(1+q^n)/(1-q^n)) \\ Lambert series \\La2(q)=sum(n=1,N,q^(2*n-1)/(1-q^(2*n-1))) \\ Lambert series (odd) \\La2(q)=sum(n=1,N,q^n/(1-q^(2*n))) \\ Lambert series (odd) La2(q)=sum(n=1,S,q^(n*(n+1)/2)/(1-q^n)) \\ Lambert series (odd) La22(q)=sum(n=1,S,q^(n*(n+1))/(1-q^(2*n))) \\ == La2(q^2) T=(1+sqrt(5))/2 \\ == golden == 1.61803398874989... \\vector(10,n,1/sqrt(5)*(T^n-(-1/T)^n)) \\ == Fibs \\be = -(1-sqrt(5))/2; \\ == 1/T == 0.61803398... \\be = 2/(1+sqrt(5)); \\ == 1/T == 0.61803398... be = (sqrt(5)-1)/2; \\ == 1/T == 0.61803398... \\vector(10,n,1/sqrt(5)*(1/be^n-(-be)^n)) \\ == Fibs " ----- EVEN:"; Fe=sum(n=1,N,1.0/fibonacci(2*n)) \\ == 1.53537050883625 Fe - sqrt(5)*sum(n=1,N, T^(2*n)/(T^(4*n)-1) ) \\ == zero Fe - sqrt(5)*( La(T^(-2)) - La(T^(-4)) ) \\ even Fe - sqrt(5)*( La(be^2) - La(be^4) ) \\ == zero Fe - sqrt(5)*( La2(be^2) ) \\ == zero Fe - sqrt(5)*( La22(be) ) \\ == zero " ----- ODD:"; Fo=sum(n=1,N,1.0/fibonacci(2*n-1)) \\ == 1.82451515740692 Fo - sqrt(5)*sum(n=1,N, T^(2*n-1)/(T^(4*n-2)+1) ) \\ == zero Fo - sqrt(5)/4*( Th3(T^(-1))^2 - Th3(T^(-2))^2 ) \\ == zero Fo - sqrt(5)/4*( Th3(be)^2 - Th3(be^2)^2 ) \\ == zero Fo - sqrt(5)/4*( Th2(be^2)^2 ) \\ == zero Fo - sqrt(5)/4*( Th22(be) ) \\ == zero " ----- BOTH:"; F1=sum(n=1,N,1.0/fibonacci(n)) \\ sum 1/Fib == 3.35988566624318... F1 - (Fe+Fo) \\ == zero F1 - sqrt(5)*sum(n=1,N, 1.0/(T^n-(-1/T)^n) ) \\ == zero F1 - sqrt(5)*sum(n=1,N, T^n/(T^(2*n)-(-1)^n) ) \\ == zero F1 - sqrt(5)*(Th22(be)/4 + La22(be)) \\ == zero \\ --------------------------------