Sign up ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

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]
share|improve this question

1 Answer 1

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}]


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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.