Find minimum n :$ \prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{\sqrt{27k^3+54k^2+36k+8}}\right)}{\arctan \left( \frac{1}{\sqrt{3k+1}} \right)}>2000$ We've got to find the minimum value of n for which
$$ \prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{\sqrt{27k^3+54k^2+36k+8}}\right)}{\arctan \left( \frac{1}{\sqrt{3k+1}} \right)}>2000$$
So I just did some simplifications hoping I'd see something:
$$\prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{(3k+2)^{\frac{3}{2}}}\right)}{\arcsin \left( \frac{1}{\sqrt{3k+2}} \right)}>2000$$
I'm thinking that maybe it's something like telescopic series and am trying to bring it into those kinds of terms. Or can it be something else? \
Edit: I think I got it
the numerator is just 3 times the denominator isn't it?

It just clicked a few minutes after I posted it
We've got:
$$\prod_{k=1}^n \frac{\arcsin\left( \frac{9k+2}{(3k+2)^{\frac{3}{2}}}\right)}{\arcsin \left( \frac{1}{(3k+2)^{\frac{1}{2}}} \right)}>2000$$
let $\sin(\theta)=\frac{1}{(3k+2)^{\frac{1}{2}}}$
now consider
$$\sin(3\theta)=\frac{3}{(3k+2)^{\frac{1}{2}}}-\frac{4}{(3k+2)^{\frac{3}{2}}}$$
notice that the argument of arcsin in the numerator is
$$\frac{9k+2}{(3k+2)^{\frac{3}{2}}}=\frac{3(3k+2)}{(3k+2)^{\frac{3}{2}}}-\frac{4}{(3k+2)^{\frac{3}{2}}}$$
which is just $sin(3\theta)$
so the whole thing simplifies to
$$3^n>2000$$
$$n>6.918$$
so if the question wants the least integral value of n, then the answer becomes 7.
