Lemma 3.2.   We have

\[F_S(a, b) = \frac{\pi}{2 M(a, b)}\]

In particular,

\[F_S(1, \sqrt{2}) = \frac{\pi}{2 M(1, \sqrt{2})}.\]

Proof. It is not hard to calculate the power series expansion

\[F(k) = \frac{\pi}{2M(1+k, 1-k)} = \sum_{n=0}^{\infty} \left( \frac{(2n-1)!!}{(2n)!!} k^n \right)^2.\]

(The "!!" means double factorial.) Calculating the AM and GM means gives us

\[M(1+k, 1-k) = M(1, \sqrt{1-k^2}).\]

Hence, applying proposition 2.1 for \(b/a = \sqrt{1-k^2}\), we get

\[a F_S(a, b) = \frac{\pi}{2 M(1, b/a)} \\ F_S(a, b) = \frac{\pi}{2 M(a, b)},\]

with the last equality following from the identity \(M(ca, cb) = cM(a, b)\). \(\blacksquare\)