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A first look at it suggests the fact the limit is precisely $1$. To check that I tried Mathematica, but no output so far. Most probably it converges very slowly. Is there any way to get the limit?

Limit[1/n^n Product[(n Sqrt[n] + (n + 1) Sqrt[k])/(Sqrt[n] + Sqrt[k]), {k, 1, n}], 
  n -> Infinity]
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up vote 12 down vote accepted

I don't think the limit is $1$.

$$\begin{split} P:=&\frac{1}{n^n}\prod _{k=1}^n \frac{n\sqrt{n}+(n+1)\sqrt{k}}{\sqrt{n}+\sqrt{k}}\\ =&\exp\left( \sum_{k=1}^{n} \log\frac{\left(\frac{1}{n}+1\right) \sqrt{\frac{k}{n}}+1}{\sqrt{\frac{k}{n}}+1} \right) \end{split}$$

So when $n\to\infty$, by expanding the $\log(\cdots)$ terms againt $1/n$ around $0$, we have a Riemann sum:

$$\lim_{n\to\infty}P = \exp\left( \int_0^1 \frac{\sqrt{x}}{\sqrt{x}+1}\,\mathrm{d}x \right)$$

Exp@Integrate[Sqrt[x]/(1 + Sqrt[x]), {x, 0, 1}]


which is approximately $1.471517765$.

Comparing with the numerical result:

With[{n = 10^7}, 
 1/n^n NProduct[(n Sqrt[n] + (n + 1) Sqrt[k])/
                (Sqrt[n] + Sqrt[k]),
               {k, 1, n}]


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Indeed. Thank you. – A_math_ninja Dec 10 '13 at 13:28
@Chris'ssis You're welcome! – Silvia Dec 10 '13 at 13:32

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