Prove that if ${s_n}$ is bounded and monotonic, then $t_n =(s_1 + \cdots+ s_n)/n$ converges to the same limit as ${s_n}$ 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 $L_1$=$L_2$? 

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