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 May 18th, 2010, 09:04 PM #1 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 sequence of the average of a sequence We know that $\lim_{n \to \infty} a_n= A \in R \Rightarrow \lim_{n \to \infty} (a_{n+1}- a_n) = 0$ and $\lim_{n \to \infty} \frac{a_1+\cdots + a_n}{n}= A \in R$ What one can say reversely?
May 18th, 2010, 11:11 PM   #2
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Re: sequence of the average of a sequence

Quote:
 Originally Posted by elim We know that $\lim_{n \to \infty} a_n= A \in R \Rightarrow \lim_{n \to \infty} (a_{n+1}- a_n) = 0$ and $\lim_{n \to \infty} \frac{a_1+\cdots + a_n}{n}= A \in R$ What one can say reversely?
The converse implication does not hold.
A simple counterexample $a_n=(-1)^n$.

The convergence of arithmetic means is called Cesaro convergence - perhaps if you google for this term you could find some resources about this topic.

E.g. http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation

 May 19th, 2010, 05:09 AM #3 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: sequence of the average of a sequence Rewriting, you said: "A sequence that converges to a real value A is a Cauchy sequence with Cesàro mean A (in the limit)." This can be reversed: "A Cauchy sequence with Cesàro mean A (in the limit) converges to A." since Cauchy sequences are convergent and hence converge to the limit of their Cesàro means. As kompik pointed out, "a sequence with Cesàro mean A (in the limit) converges to A" is false. (Of course "a Cauchy sequence converges to A" is false also.)
May 19th, 2010, 03:24 PM   #4
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Re: sequence of the average of a sequence

Quote:
 Originally Posted by kompik The converse implication does not hold. A simple counterexample $a_n=(-1)^n$. The convergence of arithmetic means is called Cesaro convergence - perhaps if you google for this term you could find some resources about this topic. E.g. http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation
But then $\lim_{n \to \infty} (a_{n+1}-a_n) \ne 0$

 May 19th, 2010, 03:29 PM #5 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 Re: sequence of the average of a sequence My Questions is: Prove or disprove that $(\lim_{n \to \infty} (a_{n+1}-a_n)= 0) \wedge ( \lim_{n \to \infty} \frac{a_1+\cdots + a_n}{n} = A \in R) \Rightarrow (\lim_{n \to \infty} a_n = A \in R)$
 May 19th, 2010, 04:21 PM #6 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: sequence of the average of a sequence Did you miss my post?
May 19th, 2010, 09:18 PM   #7
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Re: sequence of the average of a sequence

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 Originally Posted by CRGreathouse Did you miss my post?
I like to see a counterexample. Also note that $\lim_{n \to \infty} (a_{n+1}-a_n)=0$ does not mean $\left\{a_n\right\}$ is a Cauchy sequence

May 20th, 2010, 12:45 AM   #8
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Re: sequence of the average of a sequence

Quote:
 Originally Posted by elim My Questions is: Prove or disprove that $(\lim_{n \to \infty} (a_{n+1}-a_n)= 0) \wedge ( \lim_{n \to \infty} \frac{a_1+\cdots + a_n}{n} = A \in R) \Rightarrow (\lim_{n \to \infty} a_n = A \in R)$
My guess for a counterexample: take for $a_n$ the partial sums of:
$1-1+\frac12+\frac12-\frac12-\frac12+\frac13+\frac13+\frac13-\frac13-\frac13-\frac13+\dots$.

I did not try to prove this rigorously. (Perhaps someone will come up with a simpler counterexample.)
But as soon as you notice this equation expressing the sum of some consequent values of a_n:
$\frac1k+\dots+\frac{k-1}k+1+1+\frac{k-1}k+\dots+\frac1k=k$,
you can see that in the Cesaro mean you get something like
$\frac{1-2+3+\dots+(-1)^{n-1}n}{2(1+2+3+\dots+n)}$, which converges to zero.

Aside note:
Theorems like this (when Cesaro convergence + some additional conditions imply the usual convergence) are called Tauberian theorems and they are studied in summability theory.
See, for instance,
[Korevaar] Tauberian theory\ p.12,Theorem 6.1
[Montgomery Vaughan] Multiplicative Number Theory I\ p.157,Theorem 5.6
[Boos] Classical and Modern Methods in Summability Section 4.1

However, in this case they are formulated for series rather then for limits. (Cesaro summability of a series means that the Cesaro mean of partial sums converges.) Therefore we have to reformulate them slightly to our situation. The above results in our situation say that convergence of Cesaro mean implies convergence if some of the following conditions hold:
$a_{n+1}-a_n\ge0$ for each n;
$n(a_{n+1}-a_n)\to0$ for $n\to\infty$;
$n(a_{n+1}-a_n)\ge -A$ for each n; (A is some fixed constant).
There are also some results on slowly oscillating series in Boos; however, this is completely new to me, so it would take some time for me to read this.

 May 23rd, 2010, 12:23 PM #9 Senior Member   Joined: Dec 2006 Posts: 166 Thanks: 3 Re: sequence of the average of a sequence Thanks a lot kompik. Great counterexample.

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