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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?

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  • $\begingroup$ @JackD'Aurizio I don't think this example works, the Cesaro means of this sequence have $\pm 1/2$ as accumulation points. $\endgroup$ – Wojowu Oct 19 '16 at 20:19
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    $\begingroup$ @Wojowu: oh, you're right. Well, minor fix: $a_n=(-1)^n \sqrt{n}$ works (actually, $a_n=(-1)^n n^{1-\varepsilon}$ for some $\varepsilon>0$ works as well). $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 20:23
  • $\begingroup$ Oops, don't mind me missing the unbounded requirement. $\endgroup$ – Simply Beautiful Art Oct 19 '16 at 20:27
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    $\begingroup$ In general, $a_n$ must be alternating in some sense for convergence to be possible. $\endgroup$ – Simply Beautiful Art Oct 19 '16 at 20:27
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    $\begingroup$ Let's try $a_n = (-n)^n.$ If $n$ is even, then $$a_1 + \cdots + a_n >n^n - \sum_{k=1}^{n-1}k^k > n^n - \sum_{k=1}^{n-1}(n-1)^{n-1} = n^n - (n-1)\cdot (n-1)^{n-1} = n^n(1 - [(n-1)/n]^n).$$ Now $[(n-1)/n]^n \to 1/e.$ Thus for large $n$ we have $$a_1 + \cdots + a_n > n^n(1 - 1/2).$$ Dividing that by $n$ gives rapid divergence to $\infty.$ $\endgroup$ – zhw. Oct 20 '16 at 0:17
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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 $\to 0.$

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  • $\begingroup$ I think this answer deserves to be chosen as the best answer over mine! $\endgroup$ – String Oct 20 '16 at 13:10
  • $\begingroup$ @AlexQ: Why not chose this answer as the correct one? $\endgroup$ – String Oct 20 '16 at 13:11
  • $\begingroup$ Clear and precise $\endgroup$ – Alex Q Oct 20 '16 at 14:53
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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.


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.


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

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  • $\begingroup$ you could also let $a_n$ be what I had, and set $b_n=1$ for all $n.$ Then $c_n = a_n+b_n$ is unbounded and the C-means of $c_n$ tend to $1.$ $\endgroup$ – zhw. Oct 19 '16 at 23:56
  • $\begingroup$ @zhw: Yes, you are absolutely right of course :o) My aim was to make the non-zero elements "denser" to see how that could lead to non-zero limits. Your solution is very nice BTW! $\endgroup$ – String Oct 20 '16 at 0:00
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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 $0$.

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  • $\begingroup$ That is far more intuitive than my sequence, how would you prove that (vn) converges to 0 (which seems logical)? $\endgroup$ – Alex Q Oct 19 '16 at 20:32
  • $\begingroup$ the two susequences $v_{2n}$ and $v_{2n+1}$ tend to $0$, so ... $\endgroup$ – hamam_Abdallah Oct 19 '16 at 20:40

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