$\lim b_n=\frac{n^n}{(n+1)(n+2)\dots(n+n)}.$ $$b_n=\frac{n^n}{(n+1)(n+2)\dots(n+n)}.$$
Now, there is this theorem for sequences that if $\lim_{n\to ∞}⁡ a_{n+1} /a_n =l$, $|l|<1$ then $\lim_{n\to ∞}⁡ a_n=0$.
so, $\lim_{n\to ∞}⁡ b_{n+1} /b_n =e/4$ which is less than $1$, so $\lim_{n\to ∞}⁡ b_n$ should be equal to zero.
But if I calculate the limit of $b_n$ as, $b_n=(n\cdot n\cdot n\cdots n)/((n+1)(n+2)...(n+n))$ I get $\lim_{n\to ∞}⁡ b_n=1/2$.
Something is definitely going wrong. Can someone point out my mistake,Please.

One approach:
$$b_n=\frac{n^n}{(n+1)(n+2)\dots(n+n)}=\prod_{i=1}^n\frac{n}{n+i}$$
$$\implies \log b_n = n \cdot\left[ \frac{1}{n}\sum_{i=1}^n f\left(\frac{i}{n}\right)\right] \text{, where $f(x)=-\log(1+x)$}$$
So, the bracketed expression ought to tend to $\int_0^1f(x)\,dx=1-2\log2<0$
As such, $\log b_n = (1-2\log2)(n+o(1))\implies b_n=\left(\frac{e}{4}\right)^{n+o(1)}$
