Let $a_n$ be a sequence, $s_n=\sum_{i=1}^n a_i$ be partial sum.

(1) If $s_n$ is bounded, $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, show that $\lim_{n\to\infty}a_n=0$.

(2) If $\lim_{n\to\infty}s_n/n=0$, $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, do we have the statement $\lim_{n\to\infty}a_n=0$? If so, prove it, if not, construct a counterexample.

The first statement, I have no idea, just $a_n=a_1+\sum_{i=1}^{n-1}(a_{i+1}-a_i)$.

The second statement is wrong, I think; however, I could not provide a counterexample.

  • $\begingroup$ You can prove from $\lim_{n\to\infty}(a_{n+1}-a_n)=0$ that $\{a_n\}$ is a Cauchy sequence, from there, show that its limit is zero by contradiction: if that were not the case, $s_n$ wouldn't be bounded (I think I'm assuming positive terms in my last claim). $\endgroup$ – Daniel Jan 7 '15 at 1:22
  • $\begingroup$ @Devilathor How can we show that $a_n$ is Cauchy? and where $s_n$ is bounded used? $\endgroup$ – xldd Jan 7 '15 at 1:41

Here goes the proof of (1):

Suppose that $\lim_{n\to\infty}a_n=0$ doesn't hold, then, there exists $\varepsilon_0>0$ and $N\in \mathbb{N}$ such that for all $n\geq N$, $|a_n|\geq \varepsilon_0$. Here, we distinguish two cases:

CASE I: There are infinitely many negative and non-negative terms.

Then, by the condition $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, there exists $N_0$ such that for all $n\geq N_0$: $|a_{n+1}-a_n|<\varepsilon_0$, then pick some $n\geq \max\{{N_0,N}\}$ such that $a_n$ and $a_{n+1}$ have different sign, then $|a_{n+1}-a_n|\geq 2\varepsilon_0$ and $|a_{n+1}-a_n|< \varepsilon_0$, which is absurd.

CASE II: Case I doesn't hold.

Then, w.l.o.g., suppose that $M\in \mathbb{N}$ is such that $a_n$ is non-negative for all $n\geq M$, then, for all $n\geq \max\{{M,N}\}$: $a_n\geq \varepsilon_0$. This directly implies that $s_n$ can not be bounded.

This contradiction shows that $\lim_{n\to\infty}a_n=0$.

About (2), I had the idea $a_n= 1+1/n^2$, for this sequence we have $$\lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right)=0 \ \ ,$$ $$\frac{s_n}{n} = \frac{n+\sum^n_1 \frac{1}{k^2}}{n}$$ The last expression doesn't go to zero, but, it's a nice try, maybe you can fix it.

  • $\begingroup$ You were on the right track. For Case II consider $a_n = 1/n $ which is a divergent series. $\endgroup$ – Mary Jan 7 '15 at 1:55
  • $\begingroup$ @Mary No, that sequence won't work since $\lim_{n\to\infty}a_n=0$, which is what I'm trying to contradict. $\endgroup$ – Daniel Jan 7 '15 at 1:59
  • $\begingroup$ That sequence doesn't work for that neither, the fact that the serie $\sum 1/n$ is divergent implies that $s_n$ is not bounded. $\endgroup$ – Daniel Jan 7 '15 at 2:08
  • $\begingroup$ @Devilathor Suppose that $\lim_{n\to\infty}a_n=0$ does not hold, then what we could say is only that $\exists\ \epsilon_0>0$ such that $\forall\ N,\ \exists\ n\geq N$ such that $|a_n|\geq \epsilon_0$. thus, only a subsequence $a_{n_k}$ could verify $|a_{n_k}|\geq \epsilon_0$. $\endgroup$ – xldd Jan 7 '15 at 2:18
  • $\begingroup$ Ohh Yeah, you're right, I've negated wrong. Well, I'm sorry for the BIG mistake. $\endgroup$ – Daniel Jan 7 '15 at 2:20

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