Prove that if a series of complex numbers $\displaystyle\sum_{n\in{\bf Z}_{\ge0}}c_{n}$ converges to $s$ then we have $\displaystyle\sum_{n\in{\bf Z}_{\ge0}}c_{n}$ is Cesàro summable to $s$ .

My attempt:

Define $s_{j}=\displaystyle\sum_{n=0}^{j}c_{n}$ for all $j\in{\bf N}_{\ge 0}$ , then by assumption to give $\displaystyle\lim_{j\rightarrow\infty}s_{j}=s.$ Hence , given $\varepsilon>0$ , we find some integer $N_{1}>0$ such that $|s_{j}-s|<\displaystyle\varepsilon~$ provided that $j\ge N_{1}$ . Now , apply the Archimedean principle to pick a large enough $N_{2}>0$ such that $$\frac{1}{N_{2}}\sum_{j=0}^{N_{1}}|s_{j}-s|<\varepsilon$$for $N_{2}>N_{1}~.$ For each $N> N_{2} $ , one has \begin{align} \bigg|\frac{1}{N}\sum_{j=0}^{N-1}s_{j}-~s\bigg|&=\bigg|\frac{1}{N}\sum_{j=0}^{N-1}(s_{j}-\frac{N}{N}s)\bigg|\\ &\le\frac{1}{N}\sum_{j=0}^{N-1}|s_{j}-s|\\ &=\frac{1}{N}\sum_{j=0}^{N_{1}}|s_{j}-s|+\frac{1}{N}\sum_{j=N_{1}+1}^{N-1}|s_{j}-s|\\ &\le\frac{1}{N}\cdot N\varepsilon+\frac{1}{N}(N-N_{1}-1)\varepsilon\\ &<\varepsilon+\varepsilon\\ &=2\varepsilon \end{align}

Then our conclusion follows by the arbitrariness of $\varepsilon>0$ .

If my working is valid then I obtain a consequence that there is no difference of the prove between real numbers and complex numbers , basically . Did I mistake something ? Any comment or valuable suggestion I will be grateful .

  • 1
    $\begingroup$ Your argument seems valid. And yes, there is no difference between real numbers and complex numbers. The same proof works in $\mathbb{R}^d$ for any $d \geq 1$. $\endgroup$ – Sangchul Lee Mar 26 '18 at 14:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.