${\lim }_{n\to \infty }{\prod }_{k=1}^n(1+\frac{1}{k^2})$

Solution

1. Recall the Weierstrass product formula for the sine function: $\frac{\sin (\pi z)}{\pi z}={\prod }_{k=1}^{\infty }(1-\frac{z^2}{k^2})$.
2. Substitute $z=i$ into the formula to get $\frac{\sin (i\pi )}{i\pi }={\prod }_{k=1}^{\infty }(1+\frac{1}{k^2})$.
3. Apply the identity $\sin (i\pi )=i\sinh (\pi )$ to simplify the expression to $\frac{i\sinh (\pi )}{i\pi }=\frac{\sinh (\pi )}{\pi }$.
4. Hence the answer is (A).