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$ [closed]

Question

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?

Answer 1

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.