Evaluate of: ${\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$ How do we evaluate this infinite product with a sum within it?
$$\large{\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$$
Where $F_j$ is the Fibonacci number
If I open the product, it does not help me. I am sure there must be an equivalent form of this $1+\frac{1}{\sum_{j=1}^{n}F_j^2}$ into an easier manageable form.
Due to lack of knowledge in this field, I can not do much. 

We can rewrite as (due to a hint)
$${\prod_{n=1}^{\infty}\left(1+\frac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}}$$

We use Cassini's identity:
    $$F_{n-1}F_{n+1}-F_n^2=(-1)^n\Rightarrow F_{2n-1}F_{2n+1}-F_{2n}^2=1$$
    Then
    $$\prod_{n=1}^\infty \left( 1+\dfrac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}=\left( \prod_{n=1}^\infty \left( 1+\dfrac{1}{F_{2n}F_{2n+1}}\right) \right)^2$$
    Let $P_n$ be
    $$P_n=\prod_{k=1}^n \left( 1+\dfrac{1}{F_{2k}F_{2k+1}}\right),\quad P_1=1+\dfrac{1}{F_2F_3}=\dfrac{F_4}{F_3}$$
    As
    \begin{align*}
    F_{2n}F_{2n+1}+1 &= F_{2n}F_{2n+1}+F_{2n-1}F_{2n+1}-F_{2n}^2\\
    &= F_{2n+1}(F_{2n}+F_{2n-1})-F_{2n}^2\\
    &= F_{2n+1}^2-F_{2n}^2=(F_{2n+1}+F_{2n})(F_{2n+1}-F_{2n})\\
    &= F_{2n+2}F_{2n-1}
\end{align*}
    and
    $$P_2=P_1\cdot \left( 1+\dfrac{1}{F_4F_5}\right) =\dfrac{F_4}{F_3}\cdot \left( \dfrac{F_4F_5+1}{F_4F_5}\right) =\dfrac{F_4}{F_3}\cdot \dfrac{F_6\cdot F_3}{F_4F_5}=\dfrac{F_6}{F_5}$$
    We suppose that $P_n=\dfrac{F_{2(n+1)}}{F_{2n+1}}$. Then,
    \begin{align*}
    P_{n+1} &= P_n\cdot \left( 1+\dfrac{1}{F_{2(n+1)}F_{2(n+1)+1}}\right) \\
    &= \dfrac{F_{2(n+1)}}{F_{2n+1}}\left( \dfrac{F_{2(n+1)}F_{2(n+1)+1}+1}{F_{2(n+1)}F_{2(n+1)+1}}\right) \\
    &= \dfrac{F_{2(n+1)}}{F_{2n+1}}\left( \dfrac{F_{2(n+2)}F_{2n+1}}{F_{2(n+1)}F_{2(n+1)+1}}\right) \\
    &= \dfrac{F_{2(n+2)}}{F_{2(n+1)+1}}
\end{align*}
    Finally,
    $$\lim_{n\to \infty} P_n^2=\left( \lim_{n\to \infty} \dfrac{F_{2(n+1)}}{F_{2(n+1)+1}}\right)^2 =(\varphi)^2$$
