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)
$${\prod_{n=1}^{\infty}\left(1+\frac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}}$$
 A: Note that by some comments, we can first rewrite $\sum_{j=1}^nF_j^2$ as $F_nF_{n+1}$, and then rewrite the product $\displaystyle\prod_{i=1}^\infty\left(1+\frac1{F_iF_{i+1}}\right)^{(-1)^i+1}$ as $\displaystyle\prod_{i=1}^\infty \left(1+\frac1{F_{2i}F_{2i+1}}\right)^2$. But now by looking at partial products we can see that $\displaystyle\prod_{i=1}^n\left(1+\frac1{F_{2i}F_{2i+1}}\right)=\frac{F_{2n+2}}{F_{2n+1}}$ (and this can then be proven by induction), and our product is just the square of this; letting $n\to\infty$, we get the value of the product as $\phi^2=1+\phi$.
A: First, we can write the internal sum as a telescoping series
$$
\begin{align}
\sum_{k=1}^nF_k^2
&=\sum_{k=1}^nF_k(F_{k+1}-F_{k-1})\\
&=\sum_{k=1}^n(F_{k+1}F_k-F_kF_{k-1})\\[6pt]
&=F_{n+1}F_n\tag1
\end{align}
$$
Define
$$
\begin{align}
P_n
&=F_{n+2}F_{n+1}-F_{n+3}F_n\\
&=F_{n+2}F_{n+1}-(F_{n+2}+F_{n+1})F_n\\
&=F_{n+2}(F_{n+1}-F_n)-F_{n+1}F_n\\
&=F_{n+2}F_{n-1}-F_{n+1}F_n\\
&=-P_{n-1}\\
&=(-1)^n\tag2
\end{align}
$$
since $P_0=1$.
Finally,
$$
\begin{align}
\prod_{n=1}^\infty\left(1+\frac1{F_{n+1}F_{n}}\right)^{(-1)^n+1}
&=\prod_{n=1}^\infty\left(1+\frac1{F_{2n+1}F_{2n}}\right)^2\tag3\\
&=\prod_{n=1}^\infty\left(\frac{F_{2n+2}F_{2n-1}}{F_{2n+1}F_{2n}}\right)^2\tag4\\
&=\lim_{m\to\infty}\left(\frac{F_{2m+2}F_1}{F_{2m+1}F_2}\right)^2\tag5\\[6pt]
&=\phi^2\tag6
\end{align}
$$
Explanation:
$(3)$: $(-1)^n+1$ is $0$ for odd $n$ and $2$ for even $n$
$(4)$: apply $(2)$
$(5)$: write the telescoping product as the limit of the partial products
$(6)$: $\lim\limits_{n\to\infty}\frac{F_{n+1}}{F_n}=\phi$
A: 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,
    $$\lim_{n\to \infty} P_n^2=\left( \lim_{n\to \infty} \dfrac{F_{2(n+1)}}{F_{2(n+1)+1}}\right)^2 =(\varphi)^2$$
