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Let $a_i>0$ be a sequence in $\mathbb{R}$. It's well known that:

$\sum\limits_{i=0}^{n}a_i\to a \ $(as $n\to\infty)\Longrightarrow a_i\to0$ (as $n\to\infty$)

My question is when is the following statement true:

$\dfrac1n\sum\limits_{i=0}^{n}a_i\to 0\ $(as $n\to\infty$) $\Longrightarrow$ $a_i\to0$(as $n\to\infty$)

?

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  • $\begingroup$ Note that $$\sum_{i=0}^n a_i \to 0 \qquad (n\to\infty), a_i > 0$$ Is unsatisfiable, so you can deduce anything from that... $\endgroup$ – AlexR Jan 24 '14 at 12:14
  • $\begingroup$ Are the $a_i$ supposed to be independent of $n$ or should they be denoted as $a_{i,n}$? $\endgroup$ – TerranDrop Jan 24 '14 at 12:16
  • $\begingroup$ @user48805: $a_i$ is independent of $n$. $\endgroup$ – mac Jan 24 '14 at 12:21
  • $\begingroup$ @AlexR Probably the first well-known part should read $\sum\limits_{i=0}^{n}a_i\to a$(as $n\to\infty)\Longrightarrow a_i\to0$ (as $n\to\infty$) where $a<\infty$ is some finite limit. $\endgroup$ – Jeppe Stig Nielsen Jan 24 '14 at 12:25
  • $\begingroup$ @JeppeStigNielsen I guess so as well, but this would definately need to be fixed (same for the second part). $\endgroup$ – AlexR Jan 24 '14 at 12:26
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ADD This is from Apostol's "Mathematical Analysis". The work is for series, but you can aapt it for sequences

Let $s_n=\sum_{k=1}^n a_k$, $t_=\sum_{k=1}^n ka_k$, $\sigma_n=\frac 1 n\sum_{k=1}^n s_k$. Note $t_n=(n+1)s_n-n\sigma_n$.

Claim If $\sum a_n$ is $(C,1)$ summable (i.e. $\sigma_n$ converges), then $\sum a_n$ converges if and only if $t_n=o(n)$.


The converse is true, and it is a celebrated theorem of Cesàro. A counterexample to your claim would be $a_{n^2}=1$ and $a_n=2^{-n}$ otherwise.

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  • $\begingroup$ @AlexR Ah, ran past that one. Let me fix it. $\endgroup$ – Pedro Tamaroff Jan 24 '14 at 12:15
  • $\begingroup$ Possible fix: $a_n=1/2^n$ otherwise (keeping $a_{n^2}=1$). $\endgroup$ – coffeemath Jan 24 '14 at 12:23
  • $\begingroup$ @coffeemath Darn it. Positive. $\endgroup$ – Pedro Tamaroff Jan 24 '14 at 12:25
  • $\begingroup$ @PedroTamaroff: I do know it's not true in general.But My question is: under which conditions we can conclude that statement! $\endgroup$ – mac Jan 24 '14 at 12:25
  • $\begingroup$ @mac I added something. $\endgroup$ – Pedro Tamaroff Jan 24 '14 at 12:32
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A Tauberian theorem which can be deduced from Cesàro's theorem says that if the sums $$\sigma_n := \frac1n \sum_{i=1}^n \sum_{j=1}^i a_j$$ converge and $\limsup_{n\to\infty} na_n <\infty$, then the partial sums $$\sum_{i=1}^n a_i$$ converge and the limits coincide.
Thus, under the assumption $\limsup_{n\to\infty} na_n <\infty$, you could get $a_n\to0$ from $\sigma_n\to a$, but the assumption is already stronger than $a_n\to0$ so it's quite pointless.

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  • $\begingroup$ In general I know that If $n(a_{n+1}-a_n)\to 0$ then we can conclude $a_n\to a$ from $\sigma_n\to a$. But this assumption is strong! $\endgroup$ – mac Jan 24 '14 at 12:52
  • $\begingroup$ One can state something slightly more general: we have the Littlewood version of Tauber's theorem, which states the following: if a sequence of numbers $a_n$ is Cesaro summable to a number $s$, and if $a_n = O(1/n)$ (which we mean that there exists some number $M$ such that $|na_n| \leq M$ for all $n$), then the partial sums $\sum_{i = 1}^n a_n$ converges also to $s$. Unfortunately, even in this case the assumption $a_n = O(1/n)$ presupposes $a_n \to 0$, so we have rather nothing to prove. $\endgroup$ – Willie Wong Jan 24 '14 at 12:52
  • $\begingroup$ @WillieWong Just as you stated in the comment, the prerequisites are stonger than the deduction; nonetheless it answers the question ;) thanks for pointing out the erroneous definition of $\sigma_n$ $\endgroup$ – AlexR Jan 24 '14 at 12:54

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