How to prove $\lim_{n\to\infty}\frac{\log_{2}{(f_{n})}}{(\log_{2}{n})^2}=\frac{1}{2}$? For any $n\in N$, such $f_{1}=1$, and such
$$f_{2n+1}=f_{2n}=f_{2n-1}+f_{n},$$
prove that
$$\lim_{n\to\infty}\dfrac{\log_{2}{(f_{n})}}{(\log_{2}{n})^2}=\dfrac{1}{2}.$$
 A: Define $\left(~\mbox{with}\quad z \in {\mathbb C}
               \quad\mbox{and}\quad
               \left\vert z \right\vert < 1~\right)$
$$
\Psi\left(z\right)
\equiv
\sum_{n = 0}^{\infty}z^{n}\,{\rm f}_{n + 1}
=
\sum_{\sigma = \mp}\Psi_{\pm}\left(z\right)
\quad\mbox{where}\quad\left\vert%
\begin{array}{rcl}
\Psi_{-}\left(z\right) & \equiv & \sum_{n = 0}^{\infty}z^{2n + 1}\,{\rm f}_{2n + 2}
\\
\Psi_{+}\left(z\right) & \equiv & \sum_{n = 0}^{\infty}z^{2n}\,{\rm f}_{2n + 1}
\\
{\rm f}_{n}
& = &
\left.
{1 \over \left(n - 1\right)!}\,
{{\rm d}^{n - 1}\,\Psi\left(z\right) \over {\rm d}z^{n - 1}}
\right\vert_{z\ =\ 0}\,,
\quad
n > 1
\end{array}\right.
$$
\begin{eqnarray}
&&\\[5mm]
\Psi_{-}\left(z\right)
& = &
\sum_{n = 0}^{\infty}z^{2n + 1}\,{\rm f}_{2n + 2}
=
\sum_{n = 0}^{\infty}z^{2n + 1}\left\lbrack{\rm f}_{2n + 1} + {\rm f}_{n + 1}\right\rbrack
=
z\left\lbrack\Psi_{+}\left(z\right) + \Psi\left(z^{2}\right)\right\rbrack
\\
\Psi_{+}\left(z\right)
& = &
{\rm f}_{1} + \sum_{n = 0}^{\infty}z^{2n + 2}\,{\rm f}_{2n + 3}
=
1 + \sum_{n = 0}^{\infty}z^{2n + 2}\,{\rm f}_{2n + 2}
=
1 + z\,\Psi_{-}\left(z\right)
\end{eqnarray}
$$
\Psi\left(z\right)
=
z\,\Psi\left(z\right) + 1 + z\,\Psi\left(z^{2}\right)
\quad\Longrightarrow\quad
\left(1 - z\right)\,\Psi\left(z\right)
=
1 + z\,\Psi\left(z^{2}\right)
$$
$$
\Psi^{\left(n\right)}\left(0\right)
=
n\,\Psi^{\left(n - 1\right)}\left(0\right)
+
\left.n\,\Psi^{\left(n - 1\right)}\left(z^{2}\right)\right\vert_{z\ =\ 0}\,,
\qquad
n \geq 1
$$
$$
\begin{array}{rclrcl}
\Psi\left(z^{2}\right)
& = &
\sum_{n = 0}^{\infty}z^{2n}\,{\rm f}_{n + 1}
,\quad&
\Psi\left(0\right) & = & {\rm f}_{1}
\\
\Psi''\left(z^{2}\right)
& = &
2\sum_{n = 1}^{\infty}n\left(2n - 1\right)z^{2n - 2}\,{\rm f}_{n + 1}
,\quad&
\Psi\left(0\right) & = & 2{\rm f}_{2}
\\
\Psi^{\left(\tt IV\right)}\left(z^{2}\right)
& = &
2\sum_{n = 2}^{\infty}n\left(2n - 1\right)\left(2n - 2\right)\left(2n - 3\right)z^{2n - 4}\,
{\rm f}_{n + 1}
,\quad&
\Psi\left(0\right) & = & 16{\rm f}_{3}
\\
&\vdots& & \vdots&
\end{array}
$$
$$
\left.\Psi^{\left(2n\right)}\left(z^{2}\right)\right\vert_{z = 0}
=
{\rm f}_{n + 1}
$$
