I am trying to makes sense of the proof to following problem:
Given: $A_n = \displaystyle \frac{\sum_{k=1}^n a_k}{n}$.
Can $\{A_n\}$ converge if $\{a_n\}$ diverges; $\forall n,a_n>0; \limsup{a_n}=\infty$.
Proof:
Lets define $a_n = \begin{cases} k & n=k^3\\ \frac{1}{k} & n \in (k^3,(k+1)^3) \end{cases}$ and $k \in N$
The sequence looks like: $\{1,\frac{1}{2},...,\frac{1}{7},2,\frac{1}{9},...,\frac{1}{26},3...\} $
Let $k$ be maximum integer such that $k^3\le n$.
Then $A_n = \displaystyle \frac{ \sum_{p=1}^n a_p}{n} \le \frac{\sum_{j=1}^k j + \sum_{m=1}^n \frac{1}{m} }{n} = \displaystyle \frac{\sum_{j=1}^k j}{n} + \displaystyle \frac{\sum_{m=1}^n \frac{1}{m}}{n}$
(1) $\lim_{n\rightarrow \infty}\left(\displaystyle \frac{\sum_{m=1}^n \frac{1}{m}}{n}\right) = 0 $ (I omit the proof here)
(2) $\displaystyle \frac{\sum_{j=1}^k j}{n} = \frac{1}{2n}k(k+1)\le \frac{k^2+k}{2k^3}=\frac{1}{2k}+\frac{1}{2k^2}$ which converges to $0$ as $k\rightarrow \infty$
By (1) and (2) and comparison property $A_n \rightarrow 0$ as $n\rightarrow \infty$
My Question: Describing informally, is ${A_n}$ converges because in (2) "rate of divergence" of $\{\sum_{j=1}^k j\}$ is "much slower" then "rate of convergence" of $\{\frac{1}{n}\}$ to $0$ which is being expressed in (2) by square power in nominator vs cubic power in denominator?
If its incorrect view how can I better get "gut feeling" of it?
Huge thanks in advance, any help very welcome.