Find the limit of $$a_n=\frac{a^n}{\prod_{i=1}^n(1+a^i)}$$ where $a \ne -1$.
What I've got so far:
$$a_n=\frac{1}{\prod_{i=1}^{n-1}(1+a^i)}-\frac{1}{\prod_{i=1}^n(1+a^i)}$$
Now, I can see $a_n\to 0$ since the products in the denominators are divergent. I guess what I'm left with is proving that the products are divergent which I don't know how to do.
Let $$F_n(a):=\frac{a^n}{\prod_{i=1}^n(1+a^i)}.$$ Note that if $0<|a|<1 $ then $$0<|F_n(a)|\leq \frac{|a|^n}{\prod_{i=1}^n(1-|a|^i)} \leq \frac{|a|^n}{1-\sum_{i=1}^n|a|^i} =|a|^n\left(1-\frac{|a|(1-|a|^{n+1})}{1-|a|}\right)^{-1}\to 0$$ as $n$ goes to infinity. For $a=1$, $$0<F_n(1)=\frac{1}{2^n}\to 0.$$ Moreover $$F_n(1/a)=\frac{(1/a)^n}{\prod_{i=1}^n(1+(1/a)^i)}= \frac{a^{-n}\cdot a^{\sum_{i=1}^n i}}{\prod_{i=1}^n(a^i+1)}= F_n(a)\cdot a^{\frac{n^2-3n}{2}}.$$ Hence, if $|a|>1$, then $0<|1/a|<1$ and for $n\to+\infty$, $|a|^{\frac{n^2-3n}{2}}\to +\infty$, and $$|F_n(a)|=\frac{|F_n(1/a)|}{|a|^{\frac{n^2-3n}{2}}}\to 0.$$