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Let $\sum_{n=0}^\infty x_n$ be a given series of numbers,

let $S_n=\sum_{k=0}^n x_k$, $n=0,1,2,...$,

let $g\in \mathbb R$.

We say that this series is convergent to $g$ in the sense of Cesaro if $$ \frac{S_0+S_1+..+S_n}{n+1}\rightarrow g $$ as $n\rightarrow \infty$.

We say that this series is convergent to $g$ in the sense of Borel if $$ e^{-x}\sum_{n=0}^\infty S_n \frac{x^n}{n!} \rightarrow g $$ as $x\rightarrow +\infty$.

What is connection between this convergence:

  1. Is it true that if a series is convergent to some $g$ in the Cesaro sense then it is convergent to the same $g$ in the Borel sense?

  2. Is it true that if a series is convergent simultanuously in both Borel and Cesaro sense, to $g$ and $h$ respectively, then $g=h$?

Alex
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1 Answers1

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For the first question, the answer is negative. For a counterexample It suffices to take $x_n=\left\{ \begin{array}{ll} 1, & \hbox{if }n\in 2\mathbb{N} \\ 0, & \hbox{otherwise.} \end{array} \right.$.

Then $S_{2n}=S_{2n+1}=n+1,\forall n\in\mathbb{N} $

So $$\lim_{n\infty}\dfrac{S_0+\ldots+S_n}{n+1}=\frac{1}{2}$$ Which means that $(x_n)$ converges in the Cesaro sense to $\dfrac{1}{2}$.

In the other hand $$e^{-x}\sum_{n=0}^\infty S_n\frac{x^n}{n!}=e^{-x}\sum_{n=0}^\infty(n+1) \frac{x^{2n}}{(2n)!}+e^{-x}\sum_{n=0}^\infty(n+1) \frac{x^{2n+1}}{(2n+1)!}\to+\infty\mbox{ as } x\to+\infty$$

Then $(x_n)$ does not converge in the Borel sense.

Driss Alami
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  • Why the last expressions tend to infinity? – Alex May 27 '15 at 16:34
  • For $x\geq0,$$$e^{-x}\sum_{n=0}^\infty(n+1)\frac{x^{2n}}{(2n)!}\geq xe^{-x}/2\sum_{n=0}^\infty(2n)\frac{x^{2n}}{(2n)!}\geq e^{-x}/2\sum_{n=1}^\infty\frac{x^{2n-1}}{(2n-1)!}=xe^{-x}\sinh(x)\to+\infty$$ as $x\to+\infty$. We show analogously that $$e^{-x}\sum_{n=0}^\infty(n+1)\frac{x^{2n}}{(2n)!}\to+\infty$$ as $x\to+\infty$. – Driss Alami May 27 '15 at 18:55
  • Sorry, there is a typo in the first line. But it does not affect the reasonning. – Driss Alami May 27 '15 at 19:04