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 $n\to\infty$)


  • $\begingroup$ 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

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.

  • $\begingroup$ @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

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.

  • $\begingroup$ 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

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.