If $a_n\to+\infty$, $b_n\to0$, how about $\prod_{k=1}^n(1+b_ne^{-a_k})$? If $a_n$ and $b_n$ are real series, and $a_n\to+\infty$, $b_n\to0$, how is the limit
$$\underset{n\to\infty}{\lim}\prod_{k=1}^n(1+b_ne^{-a_k})?$$
There could be some specific requirements on $a_n$ and $b_n$, but I wish a more general result. Thanks~
Thanks for all the answers below, but I think I need a restriction like this:
$$a_n\sim C_1n^{\alpha},\qquad b_n\sim C_2n^{-\beta},$$
where $C_1,C_2,\alpha,\beta$ are constants, and $\alpha$, $\beta$ are positive.
I really need this problem to be solved, hope someone can help, thanks~
 A: Take $b_n = 0$, then the limit is equal to $1$.
Take $a_n = \log(\sqrt{n})$ and $b_n = \log^{-1}(n)$, then the limit is now $\infty$.
Thus, no general result exists.
A: Assuming that $b_n$ is sufficiently small, more precisely $1+b_ie^{-a_J}>0,$ we can take the logarithm and use the asymptotic $\ln (1+x)\asymp x.$ The convergence of the product is equivalent to the convergence of $c_n=b_n\sum_{k=1}^ne^{-a_k}.$ Obviously, $c_n$ does not usually converge. One may take $b_n=\frac{1}{\ln \ln{n}}$ and $a_k=\ln k.$ However, for some particular sequences one may get the results about convergence. For instance, if $b_n\asymp \frac{1}{n}$ then $c_n=\frac{\sum_{k=1}^ne^{-a_k}}{n}\to 0$ by Cesàro mean theorem. 
Edit (continuation for an updated question).
Using the conditions you mentioned above $c_n$ can be rewritten as 
$$c_n\asymp n^{-\beta}\left(\sum_{k=1}^n a^{n^{\alpha}}\right),$$
where $a=e^{-C_1}.$ Now we apply Stoltz- Cesaro theorem to compute the limit:
$$\lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{\left(\sum_{k=1}^n a^{n^{\alpha}}\right)}{n^{\beta}}=\lim_{n\to\infty}\frac{a^{(n+1)^{\alpha}}}{(n+1)^{\beta}-n^{\beta}}=\lim_{n\to\infty}\frac{a^{(n+1)^{\alpha}}}{\beta n^{\beta-1}}.$$
You can now easily analyze convergence of the last limit depending on the values of $\alpha,\beta$ and $a.$
