Let $(u_n)$ be a sequence of complex numbers that converges in mean (Cesaro convergence). Let $(v_n)$ be a sequence such that $v_n\sim u_n$. Does the sequence $(v_n)$ converge in mean?
Here is what I did. One has $v_k=u_k(1+\varepsilon_k)$ where $\lim_{k\to+\infty}\varepsilon_k=0$. So, $\begin{array}{rcl} \frac{1}{n}\sum_{k=0}^{n-1}v_k&=&\frac{1}{n}\sum_{k=0}^{n-1}u_k(1+\varepsilon_k)\\ &=&\frac{1}{n}\sum_{k=0}^{n-1}u_k+\frac{1}{n}\sum_{k=0}^{n-1}u_k\varepsilon_k. \end{array}$ I know that $\frac{1}{n}\sum_{k=0}^{n-1}u_k$ converges, say to $L\in\mathbb C$. So, now I would like to prove that $\lim_{n\to+\infty}\frac1n\sum_{k=0}^{n-1}u_k\varepsilon_k=0$. But I did not manage to do it.
Consider
$$ u_n = (-1)^n \sqrt{n} $$
and
$$ v_n = (-1)^n \sqrt{n} + \sqrt[3]{n}. $$
Note that $u_n \sim v_n$ as $n \to \infty$.
We have
$$ \left|\frac{1}{m}\sum_{n=1}^m u_n\right| \leq \frac{1}{m} \sqrt{m} = \frac{1}{\sqrt{m}}, $$
so $u_n \to 0$ in mean. But
$$ \frac{1}{m} \sum_{n=1}^{m} v_n = \frac{1}{m} \sum_{n=1}^{m} u_n + \frac{1}{m}\sum_{n=1}^{m} \sqrt[3]{n}, $$
and
$$ \sum_{n=1}^{m} \sqrt[3]{n} \sim \frac{3}{4} m^{4/3}, $$
so $\frac{1}{m} \sum_{n=1}^{m} v_n$ diverges to infinity.