The series 1-3+1-3+1-3... is (c,1) summable?

Question

To prove the series $1-3+1-3+1-3+1-3+... $ is (c,1) summable. A series $\sum_{n=1}^\infty a_n $ is said to be (c,1) summable if the sequence of partial sums $s_n$ is (c,1) summable.

A sequence ${s_n}$ is said to be (c,1) summable to L if $\lim_{n\to \infty} \sigma_n =L $

Where $\sigma_n = \frac{s_1+s_2+...+s_n}{n}$

Now for our series the sequence of partial sums is $1,-2,-1,-4,-3,-6,-5,...$

Then sequence $\sigma_n$ would be $1,-1,-2/3, -6/4, -9/5, -15/6,...$

I couldn't conclude from this that the series is (c,1) summable i think these are unwanted calculations. I think we can't apply any rearrangement since the series diverges even still we can't conclude about the convergence of $\sigma_n$ . what am i missing here?

Is there any way that i can prove the (c,1) summability of the series?

Answer 1

No, it is not (C,1) summable!

It is known that \begin{align} \liminf_n s_n\leq\liminf_n\sigma_n\leq\limsup_n\sigma_n\leq\limsup_ns_n\tag{0}\label{zero} \end{align} (see for example, Zygmund, A., Trigonometric Series, 3rd Edition, Volumes I & II, pp. 75 (Vol I), Cambridge University Press).

Notice that $s_{2n}=-n$ and $s_{2n+1}=-n+1$.

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Comments: Inequality \eqref{zero} holds in the general setting of Toeplitz transformations in which $\sigma_n=\sum^\infty_{m=1}a_{nm}s_m$ where $M=(a_{nm})$, $(n, m)\in\mathbb{N}\times\mathbb{N}$, is in infinite matrix such that

• (0) $a_{nm}\geq0$ (this is the positive condition);
• (1) for any $n\in\mathbb{N}$, $\sum^\infty_{m=1}a_{nm}<\infty$;
• (2) for any $m\in\mathbb{N}$, $\lim_{n\rightarrow\infty}a_{nm}=0$;
• (3) $\lim_n\sum^\infty_{m=1}a_{nm}=1$.

The case $a_{nm}=\frac1n\mathbb{1}(m\leq n)$ corresponds to Cesaro summability $(C,1)$.