Can an unbounded sequence have a convergent cesaro mean?

Question

I was wondering if an unbounded sequence may have a convergent cesaro mean ($\frac{1}{n}\sum_{k=1}^n a_n$). I was maybe thinking of $$a_n = (-n)^n$$ as a sequence having a convergent mean, but I might be wrong. Anyways, how would you proceed to prove such an intuition?

Answer 1

Here's an example where all $a_n$ are nonnegative. If $n = 2^m, m = 1,2, \dots ,$ define $a_n = m.$ For all other $n$ define $a_n =0.$ Then $(a_n)$ is unbounded. But if $2^m\le n < 2^{m+1},$ then

$$\frac{a_1 + \cdots + a_n}{n} \le \frac{1 + 2 + \cdots + m}{2^m} = \frac{m(m+1)/2}{2^m}.$$

The fraction on the right $\to 0$ as $m\to \infty,$ showing the Cesaro means $\to 0.$

Answer 2

Take the following example

$u_{2n}=-\sqrt{n}$

and

$u_{2n+1}=\sqrt{n}$

then

$v_{2n}=0$

and

$v_{2n+1}=\frac{\sqrt{n}}{2n+1}$.

$(u_n)$ is unbounded.

$(v_n)$ goes to $0$.

Answer 3

Here is another example following the same basic idea that zhw gave, but with non-zero limit:

For a given $n$, determine the highest power of $2$ that divides $n$, say $2^m$. Define $a_n:=m$. So we are talking about the sequence: $$ \begin{array}{|c|c|} \hline n&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&...\\ \hline a_n&0&1&0&2&0&1&0&3&0&1&0&2&0&1&0&4&...\\ \hline \end{array} $$ The idea is that $1/2$ of the sequence is at least $1$, $1/4$ is at least $2$, $1/8$ is at least $3$ etc. so on average we have $1/2+1/4+1/8+...=1$ as the Cesaro mean in the limit, as I will show in detail below.

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Then it turns out that $$ \sum_{k=1}^n a_k=\sum_{m=1}^{\lfloor\log_2 n\rfloor}\lfloor n/2^m\rfloor \approx n $$ Each term in the last sum will differ less than $1$ from the content of the floor function $$ \lfloor n/2^m\rfloor\in(n/2^m-1,n/2^m] $$ and so $$ \sum_{m=1}^{\lfloor\log_2 n\rfloor}\lfloor n/2^m\rfloor >\sum_{m=1}^{\lfloor\log_2 n\rfloor} (n/2^m-1) =\sum_{m=1}^{\lfloor\log_2 n\rfloor} n/2^m-\lfloor\log_2 n\rfloor $$ and since $$ \sum_{m=1}^t n/2^m=n-\frac{n}{2^t} $$ we have $$ \sum_{m=1}^{\lfloor\log_2 n\rfloor} n/2^m-\lfloor\log_2 n\rfloor=n-\frac{n}{2^{\lfloor\log_2 n\rfloor}}-\lfloor\log_2 n\rfloor $$ On the other hand $\sum_{m=1}^\infty n/2^m=n$ is an upper bound on this sum from which it follows that $$ n-\frac{n}{2^{\lfloor\log_2 n\rfloor}}-\lfloor\log_2 n\rfloor\leq\sum_{k=1}^n a_k\leq n $$ Hence it should be evident that $\frac1n\sum_{k=1}^n a_k\to 1$. Finally, note that $a_{2^m}=m$ so $\{a_n\}$ is unbounded.

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Multiplying the sequence from above by any $c\in\mathbb R$ we can construct a non-bounded sequence with Cesaro means converging to that $c$.