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If we have $(a_n)$ and $(b_n)$ such that $\sum a_n$ converges and $\sum b_n$ converges, I know that we do not necessarily have that $\sum c_n$ (where $(c_n)_n=a_0,b_0,a_1,b_1,\dots$ converges.

But is the Cesàro summation of $\sum c_n$ defined and if so, is it equal to $\frac12(\sum a_n+\sum b_n)$?

  • It is true that $\sum_{n=0}^\infty c_n=\sum_{n=0}^\infty a_n+\sum_{n=0}^\infty b_n$ and here's the proof. $$\sum_{n=0}^k c_n=\begin{cases}\sum_{n=0}^{\lfloor k/2\rfloor} a_n+\sum_{n=0}^{\lfloor k/2\rfloor} b_n&\text{if }k\text{ odd}\ \sum_{n=0}^{\lfloor k/2\rfloor} a_n+\sum_{n=0}^{\lfloor k/2\rfloor-1} b_n&\text{if }k\text{ even}\end{cases}$$ continues... – Sassatelli Giulio Sep 17 '23 at 10:11
  • Therefore $$\left\lvert\sum_{n=0}^k c_n-\sum_{n=0}^\infty a_n-\sum_{n=0}^\infty b_n\right\rvert\le \left\lvert\sum_{n=0}^{\lfloor k/2\rfloor} a_n-\sum_{n=0}^\infty a_n\right\rvert+\left\lvert\sum_{n=0}^{\lfloor k/2\rfloor-1} b_n-\sum_{n=0}^\infty b_n\right\rvert+\lvert b_{\lfloor k/2\rfloor}\rvert$$ and the RHS goes to zero as $k\to\infty$ because by hypothesis the differences do, and $b_n\to 0$ because its series converges. – Sassatelli Giulio Sep 17 '23 at 10:12
  • So in particular the Cesaro sum of $c_n$ is defined and it's $0$. – Sassatelli Giulio Sep 17 '23 at 10:18
  • @SassatelliGiulio I will study this proof, thank you –  Sep 17 '23 at 12:48

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