$$a_n=\prod_{i=1}^{n}\frac{16i^2-1}{16i^2-4}$$ I tried doing $\frac{a_{n+1}}{a_n}>1$ to show that the seqeunce is monotonically increasing. (For using the monotone convergence theorem)
But I cannot go ahead, to show that the sequence is bounded above too. Please help.
On
Hint: $$ \frac {16j^2 - 1} {16j^2 - 4} = 1 + \frac 3{16j^2 - 4}. $$ And $\prod_1^\infty (1 + a_n) [a_n > 0]$ converges iff $\sum_1^\infty \log(1 + a_n)$ converges. Now use comparison theorem.
Explanation:
If $\prod_1^n (1 + a_n)[a_n > 0]$ converges, then the limit $$ \lim_N \prod_1^N (1+ a_n) = \lim_N \exp \left( \sum_1^N \log(1+a_n)\right) = \exp \left(\lim_N \sum_1^N \log(1+a_n)\right) $$ exists [the last = holds according to the continuity of $\mathrm e^x$], or say $\sum_1^\infty \log(1 + a_n)$ converges.
Comparison theorem: for series $\sum c_n, \sum b_n$ with all nonnegative terms, if $c_n \leqslant b_n$ for all sufficiently large $n$, then $\sum b_n$ converges $\implies \sum c_n$ converges; $\sum c_n$ diverges $\implies \sum b_n$ diverges. A corollary: if $c_n \sim b_n [n \to \infty]$, then $\sum c_n, \sum b_n$ converge or diverge simultaneously.
On
By the Weierstrass product for the sine function $\sin x = x\prod_{n\geq 1}\left(1-\frac{x^2}{n^2 \pi^2}\right)$ we have
$$ \prod_{n\geq 1}\frac{16n^2-1}{16n^2-4} = \prod_{n\geq 1}\left(1-\frac{1}{16n^2}\right)\left(1-\frac{1}{4n^2}\right)^{-1}=\frac{\frac{2\sqrt{2}}{\pi}}{\frac{2}{\pi}}=\color{red}{\large{\sqrt{2}}}. $$
$$1+x<e^x$$ So above product is smaller than $$\large{ e^{\large\sum_{i=1}^{n}\frac{3}{16i^2-4}}}$$ This sequence is convergent, and hence bounded, so above sequence is also bounded, and hence convergent.