# Does every Cesaro summable sequence have bounded partial sums, is the sums are all positive.

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 $$\log(k)$$.

## 1 Answer

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 $$0$$.