{(2n+1)(2n+2)...(2n+n)}^1/n is the geometric average of numbers between 2n+1 and 2n+n, which is systematically lower than the arithmetic average 2n+(n+1)/2 and is higher than its lower bound 2n+1.

So (1/n){(2n+1)(2n+2)...(2n+n)}^1/n (is it what you meant?) will end up being between 2 and 2.5.

While 1/[n{(2n+1)(2n+2)...(2n+n)}^1/n] ~ 1/(2n^2) goes to zero.

For a more precise result, Indica's solution looks good.

Write S = (1/n){ ∏(k=1,n) (2n+k) }^(1/n) so Sⁿ = ∏(k=1,n) (2+k/n)

Take logs for log(S) = (1/n)∑(k=1,n) log(2+k/n)

This is just a Riemann sum for ∫[x=2,3] log(x)dx with interval 1/n

∴ lim (n→∞) log(S) = ∫[x=2,3] log(x)dx = {xlog(x)−x}, [x=2,3] = 3log(3)−2log(2)−1 = log(27/4)−1

Exponentiate for lim (n→∞) S = 27/(4e)

Assuming you meant 1 / (n*{(2n+1)(2n+2)...(2n+n)}^1/n)

Well, realize that (2n)ⁿ < {(2n+1)(2n+2)...(2n+n)} < (3n)ⁿ [why?].

Thus 2^(1/n) * n < {(2n+1)(2n+2)...(2n+n)}^(1/n) < 3^(1/n) * n

Can you take it from there?