<s>[INST] How to find the limit of the following series based on a new operation? Lets define an operation as $$a*b=a+b+ab$$  Now I need to find the following limit $$\lim_{n\to\infty}\frac{1}{1}*\frac{1}{4}*\frac{1}{9}*\frac{1}{16}\cdots*\frac{1}{n^2}$$
Now I observed the series and found that it is nothing but the sum of the magnitude of  coefficients of the equation except for the first term $$\prod_{i=1}^{n}(x-\frac{1}{i^2})$$ Now to get the sum of coefficients I substituted the value of $x$ as $-1$ thus the answer will be  $$\prod_{i=1}^{n}(-1-\frac{1}{i^2})-1$$ After this I could not proceed. My first doubt was that for $n$ odd value will be negative , and secondly I am not able to evaluate this limit.$$\prod_{i=1}^{\infty}(1+\frac{1}{i^2})$$
 [/INST] If 
$$x_n = \frac{1}{1} * \frac{1}{4} * \frac{1}{9} * \cdots * \frac{1}{n^2}$$
then 
$$x_n = x_{n-1} * \frac{1}{n^2} = (1 + x_{n-1})\left(1 + \frac{1}{n^2}\right) - 1,$$
so 
\begin{align}1 + x_n &= (1 + x_{n-1})\left(1 + \frac{1}{n^2}\right)\\
& = (1 + x_{n-2})\left(1 + \frac{1}{(n-1)^2}\right)\left(1 + \frac{1}{n^2}\right)\\
&\quad\vdots\\
& = \prod_{j = 1}^n \left(1 + \frac{1}{j^2}\right).\\
\end{align}
Now 
$$\prod_{j = 1}^n \left(1 + \frac{1}{j^2}\right) \to \frac{\sinh \pi}{\pi}\quad \text{as}\quad n\to \infty$$
using the product formula

$$\sin \pi z = \pi z \prod_{j = 1}^\infty \left(1 - \frac{z^2}{j^2}\right)$$

and substituting $z = i$.
It follows that 
$$\lim_{n\to \infty} x_n = \frac{\sinh \pi}{\pi} - 1.$$
</s>