Prove $\displaystyle\sup_{n\ge 1} \dfrac{\sum_{k=1}^{n} a_k}{A_n} = \lim_{n\to \infty} \dfrac{\sum_{k=1}^{n} a_k}{A_n}.$

Question

Let $\{a_n\}$ and $\{A_n\}$ be two sequences of non-negative numbers. Then:

$$\displaystyle\sup_{n\ge 1} \dfrac{\sum_{k=1}^{n} a_k}{A_n} = \lim_{n\to \infty} \dfrac{\sum_{k=1}^{n} a_k}{A_n}.$$

Is this right or wrong? Can you explain for me? Thanks.

Answer 1

Take $a_1 = 1, a_k = 0$ for $k \neq 1$. Take $A_n = n$.

Then $\displaystyle\sup_{n\ge 1} \dfrac{\sum_{k=1}^{n} a_k}{A_n} = 1$, $\lim_{n\to \infty} \dfrac{\sum_{k=1}^{n} a_k}{A_n} = 0$.