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By using,"Let {$a_n$} be a sequence of real numbers and let $\alpha_n=\frac{a_1+a_2+\cdots+a_n}{n}$, for all n $\in$ N. Then we say that {$a_n$} is (C,1) summable or Cesaro convergent to A iff {$\alpha_n$} converges to A." to show that the sequence {$a_n$}, where $a_n= \begin{cases} \frac{n+1}{2}, & \text{if $n$ is odd} \\ \frac{-n}{2}, & \text{if n is even} \end{cases}$, is not (C,1) summable.

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    $\begingroup$ what is $a_n$ when $n$ is even? $\endgroup$ – SiXUlm Apr 16 '16 at 20:59
  • $\begingroup$ {$a_n$} when $n$ is even should be $\frac{-n}{2}$. Sorry still getting the hang of this format $\endgroup$ – Matt Ferro Apr 16 '16 at 21:07
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If $n$ is even, say $n = 2k$, then there will be $k$ pairs, whose difference is $\frac{1}{2}$. Thus $\alpha_{2k} = \frac{1}{4}$

If $n$ is odd, say $n = 2k+1$, there will be $k$ pairs, whose difference is $\frac{1}{2}$ plus one term $a_{2k+1}$ left. Thus $\alpha_{2k+1} = \frac{k/2+k+1}{2k+1} \to \frac{3}{4}$

$2$ different sub-sequences converge to $2$ different limits, so...

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