What is the value of $\prod_{n=1}^\infty (1+\frac{1}{n^2})$? How would I go about finding $\prod_{n=1}^\infty (1+\frac{1}{n^2})$ ? Setting $a_k = \prod_{n=1}^k (1+\frac{1}{n^2})$, the first few partial products are $a_1 = 2$, $a_1 = 2.5$, $a_3 \approx 2.78$, $a_4 \approx 2.591$, $a_{20} \approx 3.50$, and they eventually appear to converge at around $3.639$. (These values were found with a quick python program. So much for my initial conjecture that the product converged to $e$.) 
I tried writing $1+\frac{1}{n^2}$ as $\frac{n^2 + 1}{n^2}$, but it wasn't immediately obvious how to prove convergence / find the limit because the product doesn't really factor or telescope. Is there a way to calculate the limit? 
 A: Since @achille hui gave an answer to the original question, let me introduce a general way of computing the product of the form
$$ \prod_{n=1}^{\infty} \frac{(n-\alpha_1)\cdots(n-\alpha_k)}{(n-\beta_1)\cdots(n-\beta_k)}. \tag{*}$$
Assumptions on parameters are as follows:


*

*$\alpha_1, \cdots, \alpha_k \in \Bbb{C}$, 

*$\beta_1, \cdots, \beta_k \in \Bbb{C}\setminus\{1,2,3,\cdots\}$,

*$\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$ (for otherwise $\text{(*)}$ either diverges or vanishes.)


Now we can write the partial product as
$$ \prod_{n=1}^{N-1} \frac{(n-\alpha_1)\cdots(n-\alpha_k)}{(n-\beta_1)\cdots(n-\beta_k)} = P \cdot\frac{\Gamma(N-\alpha_1)\cdots\Gamma(N-\alpha_k)}{\Gamma(N-\beta_1)\cdots\Gamma(N-\beta_k)} \tag{1}$$
where $P$ is defined by
$$ P = \frac{\Gamma(1-\beta_1)\cdots\Gamma(1-\beta_k)}{\Gamma(1-\alpha_1)\cdots\Gamma(1-\alpha_k)}. $$
From the Stirling's formula, we can easily check that
$$ \Gamma(N - z)
\sim \sqrt{2\pi} \, N^{N-z-\frac{1}{2}} e^{-N} \quad \text{as} \quad N \to \infty. $$
Plugging this to $\text{(1)}$ shows that $\text{(*)}$ is given by
$$ \prod_{n=1}^{\infty} \frac{(n-\alpha_1)\cdots(n-\alpha_k)}{(n-\beta_1)\cdots(n-\beta_k)} = P = \frac{\Gamma(1-\beta_1)\cdots\Gamma(1-\beta_k)}{\Gamma(1-\alpha_1)\cdots\Gamma(1-\alpha_k)}. $$
(Alternatively, one can utilize the Weierstrass' factorization of $\Gamma$ to prove this.) In our case, we can set $\alpha_1 = i$, $\alpha_2 = -i$ and $\beta_1 = \beta_2 = 0$. Then
$$ \prod_{n=1}^{\infty} \bigg( 1 + \frac{1}{n^2} \bigg) = \frac{1}{\Gamma(1+i)\Gamma(1-i)} = \frac{\sin(\pi i)}{\pi i} = \frac{\sinh \pi}{\pi}. $$
Here, we utilized the Euler's reflection formula. Similar consideration proves
$$\prod_{n=1}^{\infty} \bigg( 1 + \frac{1}{n^3} \bigg) = \frac{\cosh\left(\frac{\sqrt{3}}{2}\pi\right)}{\pi}. $$
A: Six years too late
Similar to @Felix Marin's answer to this question but focusing on the partial product
$$P_k=\prod_{n=2}^{k} \left( 1+\frac{1}{n^2}\right)=\prod_{n=2}^{k} \left( \frac{n^2+1}{n^2}\right)=\frac{\prod_{n=2}^{k}(n+i)\,\prod_{n=2}^{k}(n-i) } {\prod_{n=2}^{k}n\,\prod_{n=2}^{k}n }$$ that is to say
$$P_k=\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{\Gamma (2-i)\, \Gamma (2+i)\, \Gamma (k+1)^2}$$
But we know by Euler's reflection formula that $\Gamma (2-i)\, \Gamma (2+i)=2 \pi \, \text{csch}(\pi )$. So
$$2 \pi \, \text{csch}(\pi ) P_k=\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{ \Gamma (k+1)^2}$$ Using Stirling approximation and continuing with Taylor series
$$\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{ \Gamma (k+1)^2}=1-\frac{1}{k}+\frac{1}{k^2}-\frac{2}{3 k^3}+\frac{1}{6
   k^4}+O\left(\frac{1}{k^5}\right)$$ Taking more terms and transforming the expansion as a Padé approximant
$$\frac{\Gamma (k+(1-i))\, \Gamma (k+(1+i))}{ \Gamma (k+1)^2}=\frac {84 k^4+126 k^3+186 k^2+86 k+27 }{84 k^4+210 k^3+312 k^2+244 k+85 }+O\left(\frac{1}{k^9}\right)$$
$$P_k\sim \frac{\sinh (\pi )}{2 \pi }\,\frac {84 k^4+126 k^3+186 k^2+86 k+27 }{84 k^4+210 k^3+312 k^2+244 k+85 }$$
Using it for $k=4$
$$\frac 5 4 \times \frac{10}{9} \times \frac{17}{16} =\frac{425}{288}=1.475694444$$ while
$$P_4=\frac{32915 \sinh (\pi )}{81994 \pi }=\color{red}{1.475694}617$$
