Examine whether $\{a_n\}$ is a cauchy sequence where $a_n=\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}$ 
Examine whether $\{a_n\}$ is a cauchy sequence where $a_n=\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n}$

Attempt: If $m>n$ where $m, n$ are natural numbers,
$|a_m-a_n|=|(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2m})-(\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2n})|=|\frac{1}{2(n+1)}+\frac{1}{2(n+2)}\cdots +\frac{1}{2m}|=\frac{1}{2(n+1)}+\frac{1}{2(n+2)}\cdots +\frac{1}{2m}>\frac{1}{2(n+1)}$ 
Is the steps correct? Then what to do. Please help.

Note that
$$|a_m - a_n| = \sum_{k=n+1}^m \frac{1}{2k} > \frac{m-n}{2m} > \frac{1}{2}(1 - n/m).$$
For any $n \in \mathbb{N}$ we can choose $m > 2n$ and find
$$|a_m - a_n| > \frac1{4}.$$
Thus, it is not the case that for any $\epsilon > 0$ there exists $N$ such that $|a_m-a_n| < \epsilon$ for all $m > n > N$ 
