Let $(a_n)_{\,n\in\mathbb{N}}$ be a sequence of positive real numbers, and define $b_n = \frac{1}{n} \sum_{k=1}^n a_n$ for each $n\in\mathbb{N}$.
Suppose that consecutive elements converge:
$$\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = 1.$$
Then, intuitively, it seems plausible that $a_n$ would converge to its corresponding mean $b_n$:
$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = 1.$$
Is this limit correct? If so, how can it be shown? If not, is there a straightforward counterexample?
This problem (which is my own construction) seems closely related to the Stolz–Cesàro theorem, but my attempts to adapt the proof of that theorem have so far been unsuccessful.
Best Answer
You can't prove it because the result is not true. Consider $a_n = n$, then $a_{n + 1}/a_n = 1 + 1/n \to 1$ as $n \to \infty$, while \begin{align} \frac{a_n}{b_n} = \frac{n}{\frac{1}{2n}n(n + 1)} \to 2 \end{align} as $n \to \infty$.