Is this expression bounded?

Question

I wonder: is $$ \left( 1 + \frac{n}{a} \right)^{-a} \prod_{k = 1}^n \left( 1 + \frac{a}{k} \right) $$ uniformly bounded in $n \in \mathbb{N}$ and $0 < a \leq n$?

Following Jack's answer, I have appended the restriction $a \leq n$.

Answer 1

By the AM-GM inequality, $$\prod_{k=1}^{n}\left(1+\frac{a}{k}\right)< \left(1+\frac{a}{n}\sum_{k=1}^{n}\frac{1}{k}\right)^n<\exp\left(a H_n\right)$$ hence we just need to show that: $$ H_n - \log\left(1+\frac{n}{a}\right) = H_n - \log(n+a)+\log(a)\tag{1} $$ is uniformly bounded in $n$. This follows from $\log(n+a)>\log n$ and: $$ H_n \leq \log n +\gamma +\frac{1}{2n}. \tag{2}$$ However, for large $a$ the expression is expected to behave like $a^a$, hence we cannot have a uniform bound with respect to $a$, too. Even if $a=n$, the expression equals: $$ 2^{-n}\prod_{k=1}^{n}\frac{n+k}{k} = 2^{-n}\binom{2n}{n}=\frac{2^n}{\sqrt{\pi n}}\left(1+O\left(n^{-1/2}\right)\right).$$