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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?

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    $\begingroup$ 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 $\frac{1}{n} \sum_{k=1}^n k = \frac{1}{n}{n+1 \choose 2} \to \infty$. $\endgroup$
    – Dominik Kutek
    Commented Jul 21, 2020 at 16:59

1 Answer 1

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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 $\lim\limits_{n \to \infty} \frac 1 n \sum\limits_{k=1}^{n} u_k= \lim\limits_{n\to \infty}\frac{1}{n}\frac{(n+1)n}{2}=\infty$

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  • $\begingroup$ Can it be Cesaro summable of summable of some order? $\endgroup$
    – Anil Bagchi.
    Commented Jul 21, 2020 at 17:05
  • $\begingroup$ 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 $z\to 1$. $\endgroup$
    – alphaomega
    Commented Jul 21, 2020 at 17:31

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