A divergent sequence whose average sequence converges

It is not possible to find such a sequence. I'll do it for functions, this will give an alternative way.

Assume a locally integrable function $f$ on $[0,+\infty)$ tends to $+\infty$ at $+\infty$. We will show that the antiderivative $F(x)=\int_0^xf(t)dt$ satisfies $$ \lim_{x\rightarrow+\infty}\frac{F(x)}{x}=+\infty. $$

Fix $C>0$. Then there is $x_0>0$ such that $f(x)\geq 2C$ for all $x\geq x_0$. Thus $$ F(x)=\int_0^{x_0}f+\int_{x_0}^xf\geq 2C(x-x_0)+\int_0^{x_0}f=2Cx+D $$ hence $$ \frac{F(x)}{x}\geq 2C+\frac{D}{x}\qquad\forall x\geq x_0. $$ Now there is $x_1>x_0$ such that $D/x\geq -C$ for all $x\geq x_1$. Therefore $$ \frac{F(x)}{x}\geq 2C+\frac{D}{x}\geq 2C-C=C\qquad\forall x\geq x_1. $$ So we have show that $F(x)/x$ tends to $+\infty$ as claimed.

Now apply this to series which fulfill your assumptions. One way to do that is to construct a piecewise constant function $f$ which takes the value $a_n$ on $[n,n+1)$. Then $f(x)$ behaves like $a_n$, and $F(x)$ like the partial sums.

Sequence $(\ln \ln n)$ does not satisfy your condition. No sequence can satisfy your condition if $\lim a_n = +\infty$.

Suppose the sequence $(a_n)$ satisfies your condition and converges to positive infinity. Let $b_n = \frac{1}{n}\sum_{i=1}^n a_i$. We shall prove that $(b_n)$ is unbounded.

Let $A$ be an arbitrary positive number. There is an $n_0 \in \mathbb{N}$ such that for any $n \geq n_0$ we have $a_n > A$. Then, for any $n \geq n_0$, $$ b_n = \frac{1}{n}\left(\sum_{i=1}^{n_0}a_i + \sum_{i=n_0+1}^n a_i\right) \geq \frac{A(n-n_0)+const}{n} $$ As $n $ goes to $\infty$, the RHS converges to $A$, which means that for a sufficiently large $n$ we have $b_n > A - 1$. Since $A$ is arbitrary, sequence $(b_n)$ is unbounded and doesn't have a finite limit.

Still, if $a_n \to \infty$ means unsigned infinity, then one can build such an example. For instance: $$ a_n = (-1)^n \ln \left[\frac{n+1}{2}\right]. $$

In fact, we have the following Abelian theorem:

Theorem. If $a_n \to \alpha$ for some $\alpha \in [-\infty. \infty]$, then its average also tends to the limit $\alpha$. That is,

$$\lim_{n\to\infty} a_n = \alpha\in [-\infty. \infty] \quad\Longrightarrow\quad \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^{n} a_k = \alpha. $$

The proof for the case $\alpha = \pm\infty$ does not differ so much from that for the finite case.