# Cesàro means of conditionally convergent series

I am interested in the limit of Cesàro means (not sums or means of sums) of sequences whose corresponding series are conditionally convergent. By that I think I mean $\lim_{n \to \infty} \frac{1}{n} \sum \limits_{i=1}^n x_i$, where the sum of the terms of the sequence ($x_n$) converges conditionally. So, for example, I am interested in the limit of Cesàro means of the sequence of terms in the alternating harmonic series. Since the sequence converges to $0$, the sequence of Cesàro means also converges to $0$ (right?). But what I want to know is this. Conditionally convergent series can be rearranged to yield any value. But what about their Cesàro means? Do they stay the same after rearrangement? Or do they change?

I'm sorry if this question makes no sense.

If the partial sums of a sequence $(a_k)$ converge $$\lim_{n\to\infty}\sum_{k=1}^na_k=A$$ then its Cesàro means converge to $$\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k=\lim_{n\to\infty}\frac1n\cdot A=0$$ However, if the terms of the series are rearranged so that it diverges, we still know that for any $\epsilon\gt0$, only a finite number, $N_\epsilon$, of terms are absolutely bigger than $\epsilon$, and that finite number of terms has a finite sum, $S_\epsilon$. We then have $$\left|\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k\right|\le\lim_{n\to\infty}\left(\frac{S_\epsilon}n+\frac{n-N_\epsilon}{n}\epsilon\right)=\epsilon$$ Since $\epsilon\gt0$ was arbitrary, we have $$\lim_{n\to\infty}\frac1n\sum_{k=1}^na_k=0$$ no matter how the series is rearranged.
• @theodoricus: I apologize for the delay. I was out all day. Since the sum of the $N_\epsilon$ terms is $S_\epsilon$, there are $n-N_\epsilon$ terms whose absolute must be at most $\epsilon$. Thus, the triangle inequality says that the sum of the first $n$ terms of any rearranged sequence is at most $$\underbrace{S_\epsilon}_{\text{N_\epsilon terms}}+\underbrace{(n-N_\epsilon)\epsilon}_{\text{the rest}}$$ Then divide by $n$. – robjohn Jun 8 '15 at 5:12
• @theodoricus: Yes, it is a feature of every convergent series, conditional or not, that the terms must tend to $0$. If the terms don't tend to $0$, then the partial sums of the series will not converge – robjohn Jun 8 '15 at 9:27