Prove that if ${s_n}$ is bounded and monotonic, then $t_n =(s_1 + \cdots+ s_n)/n$ converges to the same limit as ${s_n}$

Question

I have already shown that $t_n$ is convergent using the monotonic convergence theorem. Let's say ${s_n}$ converges to $L_1$ and ${t_n}$ converges to $L_2$. How can I show that $L_1$=$L_2$?

Answer 1

Suppose $s_n \to s$ (this is true if $s_n$ are bounded and monotonic). Then $|s_n| \le B$ for some $B$.

Let $\epsilon>0$ and choose $N$ such that $|s_n - s| < {1 \over 2} \epsilon$ for all $n \ge N$. Now choose $N' \ge N$ such that ${2BN \over N'} < {1 \over 2} \epsilon$. Now suppose $n \ge N'$.

Then \begin{eqnarray} |{s_1+\cdots + s_n \over n} - s| &=& |{(s_1-s)+\cdots + (s_n-s) \over n} | \\ &\le& |{(s_1-s)+\cdots + (s_{N-1}-s) \over n} | + |{(s_{N}-s)+\cdots + (s_n-s) \over n} | \\ &\le& {2BN \over N'} + {1 \over 2} \epsilon \\ &<& \epsilon \end{eqnarray}

Answer 2

Hint: Consider

$$t_{n} - L_{1} = \frac{(s_{1} - L_{1}) + (s_{2} - L_{1}) + \dots + (s_{n_0} - L_{1})}{n} + \frac{(s_{n_0} - L_{1}) + \dots + (s_{n} - L_{1})}{n - n_{0}} \frac{n - n_0}{n}$$

Answer 3

$$s_n-t_n=s_n-\frac1n\sum_{i=1}^n s_i=s_n-\frac{ns_n-\sum_{i=1}^n s_i}n =\frac1n\sum_{i=1}^n (s_n - s_i) $$ $$|s_n-t_n|\le \frac1n\sum_{i=1}^n |s_n - s_i| = \frac1n\sum_{i=1}^N |s_n - s_i| + \frac1n\sum_{i=N+1}^n |s_n - s_i| $$

For any $\epsilon>0$ there is $N>0$, such that for every $n>N$ we have $|s_n - s_i|<\epsilon$ for each $N<i\le n$ (why?). Therefore $$0\le\lim_{n\to\infty} |s_n-t_n|\le \lim_{n\to\infty}\frac1n\sum_{i=1}^N |s_n - s_i| + \frac{n-N}n \epsilon=n\epsilon $$ Take $\epsilon\to0$ to finish the proof.