1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks very much! Two quick follow-up questions. First, could you explain what you said after "we then have..." Why do we have this? Second, is it true of every conditionally convergent series that the limit of the sequence of its terms is 0? And so, because the limit of a sequence doesn't depend on how its terms are arranged, it will always be 0? And so, because limit of the Cesàro means is the limit of the sequence, their limit will always be 0? $\endgroup$ – bakeryjake Jun 7 '15 at 16:28
  • $\begingroup$ @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 $n$. $\endgroup$ – robjohn Jun 8 '15 at 5:12
  • $\begingroup$ @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 $\endgroup$ – robjohn Jun 8 '15 at 9:27

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.