We have $ \sqrt{1+\sqrt{z}} = \sqrt{2}\left(1+\sum_{n=1}^\infty \frac{(-1)^{n-1}}{16^n n}\binom{4n - 2}{2n-1} (z-1)^n\right) $
Here is a very general method, in the particular case of an algebraic function, to prove the identity.
I — Notations
Let $f$ be the function $z \mapsto \sqrt{1+\sqrt{1+z}}/\sqrt 2$, which is holomorphic in a neighbourhood of zero. We take a determination of the square root holomorphic around 1 and such that $\sqrt 1 = 1$. Note that in order to simply the computation, I shifted the variable and normalized the value at zero.
II — Algebraic equation
The function $f$ obviously satisfies the algebraic equation $ (f(z)^2 - 1/2)^2 = \frac{1+z}{4}, $ or, equivalently, $P(f(z), z) = 0$, where $ P(Y, z) = 4Y^4 - 4Y^2 -z. $
III — Differential equation
The function $f$ satisfies the following linear differential equation : $ 16 z (z+1) f''(z) + 8(1+2z) f'(z) - f(z) = 0 $
This is a general fact that an algebraic function satisfies a linear differential equation with polynomial coefficient.
IV — Recurrence
Write $f(z) = \sum_{n\geqslant 0} u_n z^n$. Thus $ 16 z (z+1) f''(z) + 8(1+2z) f'(z) - f(z) = 16n(n-1)u_n + 16 (n+1)n u_{n+1} + 8 (n+1) u_{n+1} + 16 n u_n - u_n $ which implies that $ (4 n - 1)(4n+1)u_n + 8(n+1)(2n+1)u_{n+1} = 0. $
V — Resolution
We check easily that the sequence defined by $v_n = \frac{(-1)^{n-1}}{16^n n}\binom{4n - 2}{2n-1}$ if $n>0$ and $v_0 = 1$ satisfies the first order recurrence above. Since $u_0 = 1 = v_0$, we can conclude that $u_n = v_n$
VI — Automation
Here is a Maple session showing how to automate the proof of steps II to V.
> with(gfun): > f := sqrt(1+sqrt(1+z))/sqrt(2); 1/2 1/2 1/2 (1 + (1 + z) ) 2 f := ------------------------ 2 > holexprtodiffeq(f, y(z)); / 2 \ /d \ 2 |d | {-y(z) + (8 + 16 z) |-- y(z)| + (16 z + 16 z) |--- y(z)|, y(0) = 1} \dz / | 2 | \dz / > diffeqtorec(%, y(z), u(n)); 2 2 {(-1 + 16 n ) u(n) + (24 n + 8 + 16 n ) u(n + 1), u(0) = 1} > rsolve(%, u(n)); n GAMMA(2 n - 1/2) (-1) -1/2 ---------------------- 1/2 Pi GAMMA(2 n + 1)
You have to use the duplication formula for $\Gamma$ to retrieve the binomial coefficient.