Is the following true in general? $$\lim_{N\rightarrow \infty}\prod_{n=1}^N\left(a_n+b_n\left(\frac 1 N\right) + \mathcal{O}\left(\left(\frac 1 N\right)^2\right)\right) = \lim_{N\rightarrow \infty}\prod_{n=1}^N\left(a_n+b_n\left(\frac 1 N\right) \right)$$
If not, what are some exceptions? Are there any conditions that make the above always satisfied? ETA: One example of the above: $$e^{\frac 1 {2N}}=1+\frac 1 {2N}+\frac 1 {N^2}\frac 1 8+\frac 1 {N^3}\frac 1 {48} + \cdots $$
and
$$N\ln\left(\frac 1 {1-\frac 1 N}\right)=1+\frac 1 {2N}+\frac 1 {N^2}\frac 1 3+\frac 1 {N^3}\frac 1 4 + \cdots $$
Calculating, $\prod^N_{n=1} e^{\frac 1 {2N}} =\sqrt{e}$, it is also possible to find $$\lim_{N\rightarrow\infty}\prod^N_{n=1} N\ln\left(\frac 1 {1-\frac 1 N}\right)= \lim_{N\rightarrow\infty}\left(N\ln\left(\frac 1 {1-\frac 1 N}\right)\right)^N =\sqrt{e}$$
Solving it is very lengthy, but wolfram alpha can generate all the steps. They give the same result despite only matching on the first two terms. The compound interest definition of euler number is also an example, specifically taking $e^{\frac 1 N}$ and letting the $\mathcal{O}(N^{-2})$ function be ignored.
If both $\prod_{n=1}^N (a_n + b_n N^{-1} + \mathcal O(N^{-2}))$ and $\prod_{n=1}^N (a_n + b_n N^{-1})$ converge (and thus make sense... see the comments) then:$$\frac{\prod_{n=1}^N (a_n + b_n N^{-1} + \mathcal O(N^{-2}))}{\prod_{n=1}^N (a_n + b_n N^{-1})} = \prod_{n=1}^N\frac{ a_n + b_n N^{-1} + \mathcal O(N^{-2})}{a_n + b_n N^{-1}} = \prod_{n=1}^N(1 + \mathcal O(N^{-2}))$$ If this $\mathcal O$ notation doesn't depend on $n$ then we are done, since this is simply $(1+\mathcal O(N^{-2}))^N \longrightarrow 1$ (as $N \to \infty$).
If there is dependence on $n$ then things become trickier. As a simple example, consider: $$\prod_{n=1}^N (1 + N^{-1} + n N^{-2})= \frac{\Gamma(1+2N+N^2)}{N^{2N}\Gamma(1+N+N^2)} \longrightarrow e^{3/2}$$ $$\prod_{n=1}^N (1 + N^{-1})= (1+N^{-1})^N \longrightarrow e$$
So the $n$-dependence does really matter.