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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?

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$$\exp\sum_{k=1}^{n}\log\left(1+\frac{1}{\sqrt{k}}\right)=\exp\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k}}-\frac{1}{2k}+O\left(\frac{1}{k^{3/2}}\right)\right) $$ 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 $K\approx 3.1$.

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  • $\begingroup$ Is it possible to find an explicit value for $K$? I believe this is hopeless. $\endgroup$ – Konstantinos Gaitanas Jul 3 '18 at 16:03
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    $\begingroup$ @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. $\endgroup$ – Jack D'Aurizio Jul 3 '18 at 16:25

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