1

Given $x$, $y>0$ and being $$A_n = \left[x\cdot (x+y)\cdot (x+2y) \dots (x+(n-1)y)\right]^{1/n} $$ $$B_n = \frac{1}{n}\left[x + (x+y) + (x+2y) + \dots + (x+(n-1)y)\right] $$

I want to prove that $$\lim_{n\to\infty} \frac{A_n}{B_n}=\frac2{e}$$

I have realised that $\log A_n = \frac{1}{n}\left[\log x + \log (x+y) + \log (x+2y) + \dots + \log (x+(n-1)y)\right] $ is very similar to $B_n$, but when taking $\log \frac{A_n}{B_n} = \log A_n - \log B_n$ I don't know what to do with the term $\log B_n $. However, I think that the similarity between $\log A_n $ and $B_n$ should be helpful. Also, I know that $B_n = x+\frac{n-1}{2}y$, but I don't know if this is useful... Can someone give me a hint? Many thanks in advance!

JN_2605
  • 479

2 Answers2

2

We have that

$$A_n = \left[x\cdot (x+y)\cdot (x+2y) \dots (x+(n-1)y)\right]^{1/n}=$$

$$=\left[(n-1)!\right]^\frac1n\left[x\cdot (x+y)\cdot \left(\frac x 2+y\right) \dots \left(\frac x {n-1}+y\right)\right]^{1/n}$$

with (e.g. by Stolz-Cesaro)

$$\left[x\cdot (x+y)\cdot \left(\frac x 2+y\right) \dots \left(\frac x {n-1}+y\right)\right]^{1/n} \to y$$

and since $B_n = x+\frac{n-1}{2}y$ we reduce to evaluate

$$2\frac{\left[(n-1)!\right]^\frac1n}{n-1}$$

which is indeed $\frac 2 e$ (e.g. by Stirling).

Refer also to the related

user
  • 154,566
2

Probably too advanced.

Since you already received good answers, let me try to go beyond the limit itself.

Let $y=kx$ which makes $$x\cdot (x+y)\cdot (x+2y) \dots (x+(n-1)y)=(k x)^n\,\frac{ \Gamma \left(n+\frac{1}{k}\right)}{\Gamma \left(\frac{1}{k}\right)}$$ and so $$A_n=kx \left(\frac{\Gamma \left(n+\frac{1}{k}\right)}{\Gamma \left(\frac{1}{k}\right)}\right)^{\frac{1}{n}}$$

$$x + (x+y) + (x+2y) + \dots + (x+(n-1)y)=\frac{1}{2} n x (k (n-1)+2)$$ and so $$B_n=\frac{1}{2} x (k (n-1)+2)$$

$$\frac {A_n}{B_n}=\frac{2 k}{k (n-1)+2}\left(\frac{\Gamma \left(n+\frac{1}{k}\right)}{\Gamma \left(\frac{1}{k}\right)}\right)^{\frac{1}{n}}$$ Now, we can take logarithms and use Stirling approximation $$\log(\Gamma(p))=p (\log (p)-1)+\frac{1}{2} \log \left(\frac{2 \pi }{p}\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ and continue with Taylor series to obtain $$\log \left(\frac{A_n}{B_n}\right)=(\log (2)-1)-\frac{(k-2) \log (n)+k \log \left(\frac{\Gamma \left(\frac{1}{k}\right)^2}{2 e^2 \pi }\right)+4}{2 k n}+O\left(\frac{1}{n^3}\right)$$ which, for sure, shows the desired limit but also how it ia apprached.

Claude Leibovici
  • 260,315