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During some homeworks the following question came into my mind (it is not part of the homeworks):

Let $(a_k)_{k \in \mathbb{N}}$ be a Cesàro summable sequence in $\mathbb{C}$ and let $a := \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n a_k$. Is the sequence $$ (\sum_{k=1}^n a_k - n a)_{n \in \mathbb{N}} $$ then necessarily convergent? If not: Is it at least bounded? I tried some Cesàro summable examples and for all of them this new series converged. But I'm not sure about the general case. Is there some easy proof or some easy counterexample?

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No. Take a Cesàro summable series such that:

  • it is not summable in the standard way
  • $a = 0$

If the sequence proposed in your question were convergent, then the series will be summable in the standard way. One such example is $a_k= (-1)^k$.

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  • $\begingroup$ That was an easy counterexample, should have seen this myself...thank you! $\endgroup$ – agb Nov 28 '14 at 23:15
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Given the sequence of partial sums $$ s_n=\sum_{k=1}^na_k\tag{1} $$ the Cesàro sum is the limit of the mean of the partial sums $$ a=\lim_{n\to\infty}\frac1n\sum_{k=1}^n s_k\tag{2} $$ Therefore, $$ \lim_{n\to\infty}\left(\sum_{k=1}^na_k-na\right)=\lim_{n\to\infty}(s_n-na)\tag{3} $$ Consider the sequence $a_n=(-1)^n\sqrt{n}$ $$ \begin{align} s_{2n} &=\sum_{k=1}^{2n}(-1)^k\sqrt{k}\\ &=\sum_{k=1}^n\left(\sqrt{2k}-\sqrt{2k-1}\right)\\ &=\sum_{k=1}^n\sqrt{2k}\left(1-\sqrt{1-\frac1{2k}}\right)\\ &=\sum_{k=1}^n\sqrt{2k}\left(\frac1{4k}+\frac1{32k^2}+O\left(\frac1{k^3}\right)\right)\\ &=\frac{\sqrt{2n}}2+(2\sqrt2-1)\zeta\left(-\frac12\right)+\frac1{8\sqrt{2n}}+O\left(\frac1{n^{3/2}}\right)\tag{4}\\ s_{2n-1} &=\sum_{k=1}^{2n-1}(-1)^k\sqrt{k}\\ &=-\frac{\sqrt{2n}}2+(2\sqrt2-1)\zeta\left(-\frac12\right)+\frac1{8\sqrt{2n}}+O\left(\frac1{n^{3/2}}\right)\tag{5}\\ s_{2n-1}+s_{2n} &=(4\sqrt2-2)\zeta\left(-\frac12\right)+\frac1{4\sqrt{2n}}+O\left(\frac1{n^{3/2}}\right)\tag{6}\\ \frac1{2n}\sum_{k=1}^{2n}s_k &=(2\sqrt2-1)\zeta\left(-\frac12\right)+\frac1{4\sqrt{2n}}+O\left(\frac1{n^{3/2}}\right)\tag{7}\\ a &=(2\sqrt2-1)\zeta\left(-\frac12\right)\tag{8}\\[6pt] &\doteq-0.38010481260968401678 \end{align} $$ Thus, the Cesàro sum exists, but $$ \begin{align} \sum_{k=1}^{2n}a_k-2na &=s_{2n}-2na\\ &\sim\frac{\sqrt{2n}}2-(2n-1)(2\sqrt2-1)\zeta\left(-\frac12\right)\tag{9} \end{align} $$ which is not bounded as $n\to\infty$.

Perhaps the limit intended was $$ \lim_{n\to\infty}\left(\sum_{k=1}^ns_k-na\right)\tag{10} $$ In that case, $$ \begin{align} &\sum_{k=1}^{2n}s_k-2na\\ &=2n(2\sqrt2-1)\zeta\left(-\frac12\right)+\frac{\sqrt{2n}}4+O\left(\frac1{n^{1/2}}\right)-2n(2\sqrt2-1)\zeta\left(-\frac12\right)\\ &=\frac{\sqrt{2n}}4+O\left(\frac1{n^{1/2}}\right)\tag{11} \end{align} $$ Since $(11)$ is not bounded, $(10)$ does not exist.

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