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.