Growth of fraction of products with $\sqrt{n}$ terms

Question

Is the growth of $$f(n):=\dfrac{(n+1)(n+2)\ldots(n+\sqrt{n})}{(n-1)(n-2)\ldots(n-\sqrt{n})}$$ polynomial or not? That is, does there exist constants $k,m$ such that $$f(n)<n^k$$ for all $n>m$?

Answer 1

Yes. By considering the $i$th factor of the numerator and the $i$th last term of the denominator we get $$\frac{n+i}{n-\sqrt n-1+i} \approx \frac{n}{n-\sqrt n}\approx 1+\frac1{\sqrt n}$$ So the expression is $$f(n)\approx \left(1+\frac1{\sqrt n}\right)^{\sqrt n}\approx e $$ Especially, the sequence is $O(1)$ and therefore $f(n)<n^1$ for $n$ bi genough.

Answer 2

Since for any $0\leq k \leq n/2$ we have: $$\frac{n+k}{n-k}=\frac{1+k/n}{1-k/n}=\exp\left(2\sum_{m=0}^{+\infty}\frac{k^{2m+1}}{(2m+1)n^{2m+1}}\right)$$ it follows that $$\frac{n+k}{n-k}\in\left(\exp\frac{2k}{n},\exp\left(\frac{2k}{n}+\frac{k^2}{2n^2}\right)\right)$$ so we have: $$\prod_{k=1}^{\sqrt{n}}\frac{n+k}{n-k}\geq \exp\left(\frac{2}{n}\sum_{k=1}^{\sqrt{n}} k\right)\geq e$$ but: $$\prod_{k=1}^{\sqrt{n}}\frac{n+k}{n-k}\leq \exp\left(\frac{\sqrt{n}(\sqrt{n}+1)}{n}+\frac{1}{2n^2}\sum_{k=1}^{\sqrt{n}}k^2\right)=\exp\left(1+O\left(\frac{1}{\sqrt{n}}\right)\right),$$ so:

$$\prod_{k=1}^{\sqrt{n}}\frac{n+k}{n-k}=e\cdot\left(1+O\left(\frac{1}{\sqrt{n}}\right)\right).$$

Answer 3

$$f(n)=\frac{(n+1)(n+2)...(n+\sqrt{n})}{(n-1)(n-2)...(n-\sqrt{n})}<\frac{(n+1)(n+2)...(n+n)}{(n-1)(n-2)...(n-(n-1))}=\frac{(2n)!}{n!(n-1)!}=\frac{n(2n)!}{(n!)^2}$$ Now use Stirling's approximation for large enough $n$ $$n!\sim\sqrt{2\pi n}(\frac{n}{e})^{n}$$ Therefore $$f(n)<\frac{n(2n)!}{(n!)^2}\sim\frac{n\sqrt{2\pi 2n}(\frac{2n}{e})^{2n}}{(\sqrt{2\pi n}(\frac{n}{e})^{n})^2}=\frac{n2^{2n}}{\sqrt{\pi n}}=\frac{2^{2n}}{\sqrt{\pi}}\cdot n^{1/2}$$ This holds for all $n>1$.