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If $a_n \in \mathbb{R}$, $A_N = \sum_{n=1}^N a_n$, all partial sum so $A_N$ are positive, but their sequence oscillates and doesn't not converge, It probably down to matter but all $a_n$ are not necessarily positive, and $\lim_{K \to \infty} \frac{1}{K+1} \sum_{k=1}^K A_k \to L < \infty$ exists, Then are partial sums ${A_N}$ bounded?

Ignoring $\{a_n\}$ sequence which is a distraction, we may consider only the sequence $\{A_k\}$. This sequence has positive real numbers with a finite average. The average must grow at a rate less than $k^{\epsilon}$. So it could grow at a rate $\log(k)$.

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No, $A_{N}$ can be unbounded. Define $a_{n}$ like this:
if $n=2^{l}$ then $a_{n} = l$
if $n = 2^{l} + k, k = 1,2, ...,l$ then $a_{n} = -1$
if $n= 2^{l} + k, k = l+1, l+2, ... 2^{l+1} -1$ then $a_{n} = 0$ Note that it follows from the definition that
$A_{n} = l$ if $n=2^{l}$
$A_{n} = l-k$ if $n=2^{l} + k, k =1,2,3, ... l$
$A_{n} = 0$ if $n=2^{l} + k, k = l+1, l+2, ... 2^{l+1} - 1$
Clearly $A_{n} >=0$ and $A_{n}$ is unbounded, however Cesaro means of $A_{n}$ tend to $0$.
To see this, for given $K$ take $l$ such that $2^{l} < K <= 2^{l+1}$, we get $$\frac{1}{K+1} \sum_{n=1}^K A_n <= \frac{1}{K+1} \sum_{n=1}^{2^{l+1}} A_n = \frac{1}{K+1}\sum_{n=0}^{l}\sum_{k=2^{n}}^{2^{n+1}-1} A_{k} + \frac{1}{K+1} A_{2^{n+1}} $$ The last term tends to $0$ of course.
But the sum in each internal block, is just sum of numbers from $n$ down to $1$ so it has order of $n^{2}$. Then we sum these numbers from $n=1$ to $n=l$ so the sum will have order of $l^{3}$.
But $K$ is bigger than $2^{l}$ so the ration tends to $0$.

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