# What is the asymptotic behavior of this product?

It is not very difficult to find lower and upper bounds for the product $$\prod_{ k=1}^n \left(1+\frac{1}{\sqrt k}\right)$$
For example one can easily prove that the product is greater than $n$ and less than $2^n$ (of course these bounds can be improved).
My question is: Is it possible to determine the asymptotic behavior of this product?

## 1 Answer

$$\exp\sum_{k=1}^{n}\log\left(1+\frac{1}{\sqrt{k}}\right)=\exp\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k}}-\frac{1}{2k}+O\left(\frac{1}{k^{3/2}}\right)\right)$$ equals $$\exp\left[2\sqrt{n}-\frac{1}{2}\log(n)+O(1)\right]$$ hence your product behaves like $\frac{e^{2\sqrt{n}}}{K\sqrt{n}}$ for large values of $n$.
Are you interested in a explicit value for $K$? We approximately have $K\approx 3.1$.

• Is it possible to find an explicit value for $K$? I believe this is hopeless. – Konstantinos Gaitanas Jul 3 '18 at 16:03
• @KonstantinosGaitanas: as the exponential of a convergent series, or the exponential of a not-so-terrible integral, sure. In terms of standard mathematical constants, I believe not. – Jack D'Aurizio Jul 3 '18 at 16:25