I came across the following problems on convergence of sequences during the course of my self-study of real analysis:
Suppose $a_n \to a$. Define $s_n = \frac{1}{n}\sum_{k=1}^{n} a_k$ Prove that $s_n \to a$.
So $(a_n-a)$ is a null sequence. I want to show that $(s_n-a)$ is a null sequence. By a previous exercise, I know that $(x_n)$ is a null sequence $\implies$ $(y_n)$ is a null sequence where $y_{n} = (x_1+ \cdots+ x_n)/n$. So can we do something analogous to "adding $a$ to both sides" to get the desired result?
Show that the sequence $a_n = \left(1- \frac{1}{2} \right) \left(1- \frac{1}{3} \right) \cdots \left(1- \frac{1}{n+1} \right)$ is convergent.
So $a_1 = \frac{1}{2}$, $a_2 = \frac{1}{3}, \dots, a_n = \frac{1}{n+1}$. So I conjecture that $(a_n)$ is a null sequence. In other words, for each $\epsilon >0$, $|a_n| \leq \epsilon$ for all $n>N$. Let $\epsilon = \frac{1}{n}$. Choose $N = n+1$. Then the convergence follows?
Prove that the sequence $a_n = \frac{1}{n+1}+ \frac{1}{n+2} + \dots + \frac{1}{n+n}$ is convergent to a limit $\leq 1$.
So $a_{n} < a_{n+1}$ for all $n$. Then I need to show that it is bounded above by $1$. To show this should I consider $(1-a_n)$? All the terms are $<1$.