Let $\{a_n\}_{n=1}^{\infty}$ be a monotonic decreasing sequence of positive real numbers. Show that $\lim_{k \rightarrow \infty} \frac 1 k \sum_{n=1}^k a_n = \inf a_n$
Suppose we take the sum of the first $k =m$ numbers then the average will be greater than the smallest number. However per definition of $\inf a_n$ we know that $\inf a_n \le a$ where $a \in A = \{a_n \mid n \in \mathbb N\}$. This holds for any $m$. Can someone give a hint or explanation how to proof this ?
We will show that $\lim_{k \rightarrow \infty} \frac 1 k \sum_{n=1}^k a_n = \inf a_n=a$ and $a_n\to a$ because it is a monotonic decreasing sequence of positive real numbers.
Because $a_n\to a$ there is a $M>0:|a_n|\leq M$ for every $n\geq 1$. Just to make it more simple suppose that $a=0$(because if we set $b_n=a_n-a$ we have that $b_n\to 0$ and thus it's enough to show that $\frac {b_1+b_2+...+b_n}{n}\to 0 $ if $b_n\to 0$).
Let $ε>0$.Because $\frac {M}{\sqrt n}\to 0$ there is a $n_o\in \Bbb N:|a_n|<ε/2$ and $\frac {M}{\sqrt n}<ε/2$ for $n\geq n_0$. So for $n\geq n_0^2$ we have $|\frac {a_1+a_2+...+a_n}{n}|\leq \frac {a_1+a_2+...+a_{\sqrt n}}{n}|+|\frac {a_{\sqrt {n+1}}+...+a_n}{n}|\leq \frac {M[\sqrt n]}{n}+\frac {nε}{2}\frac {1}{n}\leq \frac {M}{\sqrt n}+\frac {ε}{2}\leq \frac {ε}{2}+\frac {ε}{2}=ε$ and thus $\frac {a_1+a_2+...+a_n}{n}\to a$ for $n\to \infty$
