Given that $x_n$ is a real sequence, $s_n = \frac{x_0+x_1+\cdots+x_n}{n+1}$ and $s_n$ converges, $a_n = x_n-x_{n-1}$, $na_n$ converges to 0, and $x_n-s_n=\frac{1}{n+1}\sum_{i=1}^n ia_i$, prove $x_n$ converges, and that $x_n$ and $s_n$ converge to the same limit.

Since I don't know anything about the sequences increasing, decreasing, or being positive, I was thinking that I can show that $\frac{1}{n+1}\sum_{i=1}^n ia_i$ converges, so then $x_n$ will be the sum of two convergent sequences and so it must converge, but I don't know how to show $\frac{1}{n+1}\sum_{i=1}^n ia_i$ converges, or if this is even the right way to go about this problem.

Also since $s_n$ converges, it is bounded so maybe I can use that fact, but I'm not sure how that could help.

I also showed that $a_n$ must converge to 0 as well using the squeeze theorem, but I don't know if that helps either.

Any hints would be great. I have been stuck on this for days.

  • $\begingroup$ @David Yes, thank you, I was wondering how to format it like that. $\endgroup$ – lee Mar 31 '14 at 3:03

Since $na_n\to 0$, Cesaro gives $$x_n-s_n=\frac{1}{n+1}\sum_{i=1}^n ia_i\to 0$$

Can you move on?

Add Recall Cesaro's theorem says that if $z_n\to \ell$; then the averages $\overline {z_n}=\displaystyle \frac 1 {n+1}\sum_{k=0}^n z_n$ also converge to $\ell$.

  • 1
    $\begingroup$ So since we know $s_n$ converges, we can do operations on limits. Let $x_n = x_n-s_n+s_n$. $\lim_{n\to \infty}x_n-s_n = 0$ and $\lim_{n\to \infty}s_n = l$, so we can rewrite this as $\lim_{n\to \infty}{x_n-s_n+s_n}$ = $\lim_{n\to \infty}x_n$ and therefore the limit of $x_n$ exists. And with $x_n-s_n$ converging to 0 we can show their limits are equal. But could you explain a little more how you got $x_n-s_n$ converging to 0? @Pedro $\endgroup$ – lee Mar 31 '14 at 3:42

First, let me expand on Pedro's answer. Suppose $na_n$ converges to 0. We claim that the sequence $\dfrac{1}{n+1}(a_1+2a_2+...+na_n)$ converges to 0. Let $\epsilon > 0$. Then there is $N$ so that for all $n>N, |na_n| < \epsilon$. Let $M = |a_1+2a_2+...+Na_N|$. Then for any $n > N$, $|a_1+2a_2+...+na_n| \leq M + |(N+1)a_{N+1}+...+na_n| < M+(n-N)\epsilon$. Therefore, whenever $n>N$, $\dfrac{1}{n+1}|a_1+2a_2+...+na_n|<\dfrac{M}{n+1}+(\dfrac{n}{n+1}-\dfrac{N}{n+1})\epsilon$. Let $N'$ be such that for all $n>N'$, $\dfrac{M}{n+1} < \dfrac{\epsilon}{2}$, so that $\dfrac{1}{n+1}|a_1+2a_2+...+na_n|<1.5\epsilon$.

So $x_n-s_n$ converges to 0.

Let $s$ be the limit of the sequence $(a_n)$.

Then $|x_n-s|=|x_n-s_n+s_n-s|\leq |x_n-s_n|+|s_n-s|$. Let $\epsilon > 0$ be given. Then for sufficiently large $n$ the term on the right is less than $\epsilon$, which shows that the sequence $(x_n)$ converges to $s$.

  • $\begingroup$ how did you get $1.5\epsilon$? And don't we just want it to be less than $\epsilon$? (not $1.5\epsilon$?) @Mehta $\endgroup$ – lee Mar 31 '14 at 6:04
  • $\begingroup$ $\dfrac{M}{n+1} < \epsilon/2$ and $\dfrac{n}{n+1}-\dfrac{N}{n+1} < \epsilon$. Also we don't need strictly less than $\epsilon$ in the proof of these types of arguments. If you want, let $\epsilon > 0$ and define $\epsilon' = \epsilon/1.5$. Then if you use $\epsilon'$ instead of $\epsilon$ in the proof, you'll get that the term is $< 1.5(\epsilon')=\epsilon$. This is sometimes useful to know when working on real analysis proofs. $\endgroup$ – Mehta Mar 31 '14 at 7:52
  • $\begingroup$ You want to assume $na_n$ goes to zero. Else things are incorrect. $\endgroup$ – Pedro Tamaroff Mar 31 '14 at 12:24
  • $\begingroup$ So we can say $\dfrac{M}{n+1} < \epsilon/2$ because $\dfrac{M}{n+1}$ converges to 0? And $(\dfrac{n-N}{n+1})\epsilon < \epsilon$ because $(\dfrac{n-N}{n+1})\epsilon$ converges to $\epsilon$? @Mehta $\endgroup$ – lee Mar 31 '14 at 20:22
  • $\begingroup$ @lee: since $\dfrac{M}{n+1}$ converges to 0, then eventually, for $n$ large enough, $\dfrac{M}{n+1}<\epsilon/2$. $\endgroup$ – Mehta Mar 31 '14 at 21:29

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