Let $a_n = a_1 + (n-1)d$ and $b_n=\frac{\sqrt[n]{a_1 \cdot a_2 \cdot\ldots \cdot a_n}}{\frac{a_1+a_2+\ldots+ a_n}{n}}$
Prove that $\lim_{n \to \infty}b_n$ exists and find it for $d>0$ and $a_1>0$.
My attempt:
Because an arithmetic average is bigger than a geometric one, then $0\le b_n \le 1$
The sum of $a_n: S_n = \frac{n(2a_1 +(n-1)d)}{2}$ (arithmetic sequence)
Then $b_n = \frac{\sqrt[n]{a_1 \cdot a_2 \cdot… \cdot a_n}}{\frac{\frac{n(2a_1 +(n-1)d)}{2}}{n}} = \frac{2\sqrt[n]{a_1 \cdot a_2 \cdot… \cdot a_n}}{2a_1 +(n-1)d}$
I tried to prove that $b_n \ge b_{n+1}$, but couldn't (then I would have been able to say that $\lim_{n \to \infty}b_n$ exists.)
Any hints would be appreciated. Thanks!
Best Answer
This is a sketch.
Set $A_n:=\prod^n_{k=1}\frac{a+b(k-1)}{a+\frac{b}{2}(n-1)}$. Then \begin{align*} \frac{A_{n+1}}{A_n}&= \frac{a+bn}{a+\frac{b}{2}n}\left(\frac{a+\frac{b}{2}(n-1)}{a+\frac{b}{2}n}\right)^n \\ &= \frac{a+bn}{a+\frac{b}{2}n}\left(1-\frac{b}{2a+bn}\right)^n\rightarrow2 e^{-1} \end{align*} The convergence of $A_{n+1}/A_n$ implies the converge of $\sqrt[n]{A_n}$ and they have the same limit.