Evaluate of: ${\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$

Question

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.

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We can rewrite as (due to a hint)

$${\prod_{n=1}^{\infty}\left(1+\frac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}}$$

Answer 1

You know that:

$$\prod_{n\ge1}^{ }\left(1+\frac{1}{\sum_{j=1}^{n}F_{j}^{2}}\right)^{\left(-1\right)^{n}+1}=\prod_{n\ge1}^{ }\left(1+\frac{1}{F_{n}F_{n+1}}\right)^{\left(-1\right)^{n}+1}$$

$$=\exp\left(\ln\left(\prod_{n\ge1}^{ }\left(1+\frac{1}{F_{n}F_{n+1}}\right)^{\left(-1\right)^{n}+1}\right)\right)=\exp\left(\sum_{n\ge1}^{ }\left(\left(-1\right)^{n}+1\right)\ln\left(1+\frac{1}{F_{n}F_{n+1}}\right)\right)$$

Using the equality $\ln\left(1+\frac{1}{x}\right)<\frac{1}{x}$ and considering the fact that $\left(-1\right)^{n}+1$ is either $2$ or $0$ follows:

$$\bbox[5px,border:2px solid #C0A000]{1=\exp\left(0\right)\le\exp\left(\sum_{n\ge1}^{ }\left(\left(-1\right)^{n}+1\right)\ln\left(1+\frac{1}{F_{n}F_{n+1}}\right)\right)}\tag{I}$$

$$\exp\left(\sum_{n\ge1}^{ }\left(\left(-1\right)^{n}+1\right)\ln\left(1+\frac{1}{F_{n}F_{n+1}}\right)\right)<\exp\left(\sum_{n\ge1}^{ }\left(\left(-1\right)^{n}+1\right)\frac{1}{F_{n}F_{n+1}}\right)$$

Split the summation into two parts:

$$\color{red}{\exp\left(\sum_{n\ge1}^{ }\frac{\left(-1\right)^{n}}{F_{n}F_{n+1}}+\sum_{n\ge1}^{ }\frac{1}{F_{n}F_{n+1}}\right)}$$

It's known that:

$$\sum_{n=k}^{mk}\frac{\left(-1\right)^{n}}{F_{n}F_{n+1}}=\frac{F_{k+1}}{F_{k}}-\frac{F_{mk+2}}{F_{mk+1}}$$

Which is true when $n\ge1\ ,\ m\ge2$.

$$\text{and}$$ $$\frac{1}{F_{k}^{2}+1}<\sum_{n=k}^{mk}\frac{1}{F_{n}F_{n+1}}<\frac{1}{F_{k}^{2}}$$

Which is true when $n$ is even.

Setting $k \mapsto 1$ yields:

$$\sum_{n=1}^{m}\frac{\left(-1\right)^{n}}{F_{n}F_{n+1}}=\frac{F_{1+1}}{F_{1}}-\frac{F_{m+2}}{F_{m+1}}=1-\frac{F_{m+2}}{F_{m+1}}\tag{1a}$$ $$\text{and}$$

Setting $k \mapsto 2$ yields:

$$\frac{1}{2}=\frac{1}{F_{2}^{2}+1}<\sum_{n=2}^{2m}\frac{1}{F_{n}F_{n+1}}<\frac{1}{F_{2}^{2}}=1\tag{1b}$$

Fibonacci numbers for $n \in \mathbb N$ form an increasing sequence, using this fact we conclude:

$$F_n \le F_{n+1}$$ Setting $n \mapsto m+1$ we have: $$F_{m+1} \le F_{m+2} \:\:\:\:\:\:\text{or equivalently}\:\:\:\:\:\: -\frac{F_{m+2}}{F_{m+1}}<-1\:\:\:\:\: $$

Note that based on the main index we can be sure that diving a term by the other one in the sequence is always well-defined.

The red part can be written as :

$$\color{red}{\exp\left(\sum_{n\ge1}^{ }\frac{\left(-1\right)^{n}}{F_{n}F_{n+1}}+\sum_{n\ge1}^{ }\frac{1}{F_{n}F_{n+1}}\right)}=\exp\left(\lim_{m\to\infty}\sum_{n=1}^{ m}\frac{\left(-1\right)^{n}}{F_{n}F_{n+1}}+\lim_{m\to\infty}\sum_{n=1}^{ 2m}\frac{1}{F_{n}F_{n+1}}\right)$$ Using $\text{(1a)}$, $\text{(1b)}$ and observing that $f(x)=e^x$ is strictly monotonic over $\mathbb R$, we have:

$$ \exp\left(\lim_{m\to\infty}\sum_{n=1}^{ m}\frac{\left(-1\right)^{n}}{F_{n}F_{n+1}}+\lim_{m\to\infty}\sum_{n=1}^{ 2m}\frac{1}{F_{n}F_{n+1}}\right)< \exp\left(\lim_{m\to\infty}1-\frac{F_{m+2}}{F_{m+1}}+1+1\right)$$

On the other hand: $$\bbox[5px,border:2px solid #C0A000]{\exp\left(\lim_{m\to\infty}1-\frac{F_{m+2}}{F_{m+1}}+1+1\right)<\exp\left(2\right)=e^2\simeq 7.38905609893}\tag{II}$$

Combining $\text{(I)}$,$\text{(II)}$ follows:

$$1\le\prod_{n\ge1}^{ }\left(1+\frac{1}{\sum_{j=1}^{n}F_{j}^{2}}\right)^{\left(-1\right)^{n}+ 1}<7.38905609893$$

The real answer is approximately $\color{blue}{2.61803398875}$.

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Another upper bound can be found using the fact that since $\left(-1\right)^{n}+1$ is either $2$ or $0$ , so the product maybe written as :

$$\prod_{n\ge1}^{ }\left(1+\frac{1}{F_{2n}F_{2n+1}}\right)^{2}$$$$=\exp\left(2\sum_{n\ge1}^{ }\ln\left(1+\frac{1}{F_{2n}F_{2n+1}}\right)\right)<\exp\left(2\sum_{n\ge1}^{ }\frac{1}{F_{2n+1}}\right)=\exp\left(2\sum_{n\ge0}^{ }\frac{1}{F_{2n+1}}-2\right)$$

Using sums of reciprocals of odd-indexed Fibonacci numbers follows: $$\simeq \exp\left(2\left(1.8245151574069245681\right)-2\right)=\bbox[5px,border:2px solid #00A000]{5.20193314322}$$

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Evaluating the limit: Rewrite your product as : $$\prod_{n\ge1}^{ }\left(1+\frac{1}{F_{2n}F_{2n+1}}\right)^{2}$$

Then use d'Ocagne's identity:

$$F_{2n+2}F_{2n+1}-\left(F_{2n+2}+F_{2n+1}\right)F_{2n}=1$$ $$F_{2n+2}\left(F_{2n+2}-F_{2n}\right)-F_{2n}F_{2n+2}=1+F_{2n}F_{2n+1}$$ $$F_{2n+2}^{2}-2F_{2n}F_{2n+2}=1+F_{2n}F_{2n+1}$$ $$F_{2n+2}\left(F_{2n}+F_{2n+1}\right)-2F_{2n}F_{2n+2}=1+F_{2n}F_{2n+1}$$ $$F_{2n+2}F_{2n+1}-F_{2n}F_{2n+2}=1+F_{2n}F_{2n+1}$$ $$F_{2n}F_{2n+2}+F_{2n-1}F_{2n+2}-F_{2n}F_{2n+2}=1+F_{2n}F_{2n+1}$$ $$\color{magenta}{F_{2n-1}F_{2n+2}=1+F_{2n}F_{2n+1}}$$ The rest of the answer has been answered by robjohn.

Answer 2

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 $\phi^2=1+\phi$.

Answer 3

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
$(6)$: $\lim\limits_{n\to\infty}\frac{F_{n+1}}{F_n}=\phi$

Answer 4

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, $$\lim_{n\to \infty} P_n^2=\left( \lim_{n\to \infty} \dfrac{F_{2(n+1)}}{F_{2(n+1)+1}}\right)^2 =(\varphi)^2$$