Convergence of $\prod\limits_{k=1}^n \left(1+\frac{1}{k^2}\right) $ How can you determine whether the limit
$$ \displaystyle \lim_{n\to\infty} \prod_{k=1}^n \left(1+\frac{1}{k^2}\right)  $$ 
exists? Thanks for any hints!  
 A: When checking convergence of infinite products of factors of the form $1 + \text{something positive}$, the inequalities
$$ 1 + \sum_{k=1}^n a_k \le \prod_{k=1}^n (1+a_k) \le \exp \sum_{k=1}^n a_k $$
are useful.  The lower one shows that if the infinite sum diverges, the infinite product does too; the upper one shows that if the infinite sum converges, the infinite product does too.  (To prove the lower inequality, imitate the proof of Bernoulli's inequality, which is a special case; to prove the upper, use $1+x\le e^x$.)
In your case, the corresponding sum is $\sum\limits_{k=1}^\infty \frac1{k^2}$, whose convergence can be checked by $p$-test, or by integral test, or by comparison to the telescoping sum $$1+\sum_{k=2}^\infty \frac1{k(k-1)} = 1+\sum_{k=2}^\infty \left(\frac1{k-1} - \frac1k\right),$$ or by recalling the result that made Euler famous, his solution to the Basel problem.
A: $$\frac{\sinh \pi x}{\pi x}=\prod_{k=1}^{\infty }\left(1+\frac{x^2}{k^2}\right)$$
is the Weierstrass product for the hyperbolic sine function, see here
or here. If you plug $x=1$:
$$\prod_{k=1}^{+\infty}\left(1+\frac{1}{k^2}\right) = \frac{\sinh\pi}{\pi}.$$
A: Since for any $|x|<1$ we have $1+x\leq\frac{1}{1-x}$, for any $n\geq 2$:
$$\prod_{k=1}^n\left(1+\frac{1}{k^2}\right)=\frac{3}{2}\prod_{k=2}^{n}\left(1+\frac{1}{k^2}\right)\leq\frac{3}{2}\prod_{k=2}^{n}\frac{k}{k-1}\cdot\frac{k}{k+1}=\frac{3n}{n+1}\leq 2.$$
