# Is the product of two Cesaro convergent series Cesaro convergent?

Let $$\{a_n \}_{n \geq 1}$$ and $$\{b_n \}_{n \geq 1}$$ be two sequences of real numbers such that the infinite series $$\sum\limits_{n=1}^{\infty} a_n$$ and $$\sum\limits_{n=1}^{\infty} b_n$$ are both convergent in the Cesaro sense i.e. \begin{align*} \lim\limits_{n \to \infty} \frac 1 n \sum\limits_{k=1}^{n} s_k & < + \infty \\ \lim\limits_{n \to \infty} \frac 1 n \sum\limits_{k=1}^{n} t_k & < + \infty \end{align*}

where $$\{s_k \}_{k \geq 1}$$ and $$\{t_k\}_{k \geq 1}$$ are sequences of partial sums of the series $$\sum\limits_{n=1}^{\infty} a_n$$ and $$\sum\limits_{n=1}^{\infty} b_n$$ respectively. Can I say that $$\sum\limits_{n=1}^{\infty} a_n b_n$$ is convergent in the Cesaro sense? If "yes" then what can I say about it's limit in terms of the limits of the given two series?

• What if $a_n=b_n = (-1)^{n+1}$. Then $s_{2k}=t_{2k} = 0$ and $s_{2k+1}=t_{2k+1}=1$, hence $\frac{1}{n} \sum_{k=1}^n s_k$ is convergent, but $a_nb_n = 1$ and its partial sums $S_k = k$, but $\frac{1}{n} \sum_{k=1}^n k = \frac{1}{n}{n+1 \choose 2} \to \infty$. – Dominik Kutek Jul 21 '20 at 16:59

No. Consider $$a_{n}=b_n=(-1)^n$$. Then both of them are Cesaro summable but $$c_n=a_n \cdot b_n= 1$$ isn't, since $$\lim\limits_{n \to \infty} \frac 1 n \sum\limits_{k=1}^{n} u_k= \lim\limits_{n\to \infty}\frac{1}{n}\frac{(n+1)n}{2}=\infty$$
• I do not know what you may have in mind. For example $c_n$ is not even Abel summable, even though $a_n$, $b_n$ are, since $(|z|<1)$, $\sum_{n=1}^{\infty}c_nz^n = \sum_{n=1}^{\infty}z^n=\frac{1}{1-z}$ which blows up for $z\to 1$. – alphaomega Jul 21 '20 at 17:31