0

enter image description here

The answer to the question is $\frac{\sqrt{2\pi}}e$, although I do not know how to come to this solution. This is a problem from a math contest from the year 2016.

  • 2
    The $N$th partial product is $$\prod_{n=1}^N\left(en^{n+1/2}(n+1)^{-(n+1)+1/2}\right)=e^N\prod_{n=1}^Nn^{n+1/2}\prod_{n=2}^{N+1}n^{-n+1/2}=e^NN!(N+1)^{-N-1/2}$$ and Stirling formula readily yields the result. – Did Jan 10 '17 at 00:47

2 Answers2

2

Hint: \begin{align*} \mathcal{I}&=\prod_{n=1}^\infty \Bigg [e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}} \,\Bigg ]=\lim_{N\to\infty}\prod_{n=1}^N \Bigg [e\left(\frac{n}{n+1}\right)^{n}\sqrt{\frac{n}{n+1}} \,\Bigg ]\\ &=\lim_{N\to\infty}e^N\left[\left(\frac 12\right)^1\left(\frac 23\right)^2\left(\frac 34\right)^3\cdots \left(\frac N{N+1}\right)^N\right] \sqrt{\frac12\cdot\frac23\cdot\frac34\cdots\frac{N}{N+1}}\\ &=\lim_{N\to\infty}e^N\left[\frac{1\cdot 2\cdot 3\cdots N}{(N+1)^N}{\cdot \frac{N+1}{N+1}}\right]\sqrt\frac1{N+1}\\ &=\lim_{N\to\infty}e^N\frac{{(N+1)!}}{(N+1)^{N+\frac{3}{2}}}\\ \end{align*} Hope you can take it from here using Stirling formula.

Did
  • 279,727
Renascence_5.
  • 5,281
  • 1
  • 27
  • 69
  • Hey I know this is old but how can I use Stirling's formula on this? I tried researching it. –  Jan 16 '17 at 04:43
  • @Ekesh $\displaystyle \left ( N+1 \right )!\sim \sqrt{2\pi(N+1)}\left(\frac{N+1}e\right)^{N+1}$, take some limit and you will get the answer. – Renascence_5. Jan 16 '17 at 04:46
0

The logarithm of such product is given by $$ \sum_{n\geq 1}\left(1-\left(n+\frac{1}{2}\right)\log\frac{n+1}{n}\right)=\sum_{n\geq 1}\int_{n}^{n+1}\frac{n+\tfrac{1}{2}-x}{x}\,dx $$ that is: $$ -\int_{-\frac{1}{2}}^{\frac{1}{2}}\sum_{n\geq 1}\frac{t}{n+\tfrac{1}{2}+t}\,dt = -\int_{0}^{1}\sum_{n\geq 1}\frac{u^2}{(2n+1)^2-u^2}\,du$$ or, by considering the logarithmic derivative of the Weierstrass product for the cosine function, $$ -\int_{0}^{1}\left(\frac{u^2}{u^2-1}+\frac{\pi u}{4}\tan\left(\frac{\pi u}{2}\right)\right)\,du $$ that is: $$ -\lim_{t\to 1^-}\left(t-\text{arctanh}(t)-\tfrac{t}{2}\log\cos\tfrac{\pi t}{2}+\tfrac{1}{2}\log 2\right){=}\tfrac{1}{2}\log(2\pi)-1. $$ Exponentiating back, we get that the value of the infinite product is $\color{red}{\large\frac{\sqrt{2\pi}}{e}}$ without exploiting Stirling's approximation.

Jack D'Aurizio
  • 353,855