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.
Best Answer
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 $s$.