Proof that a sequence is Cauchy. Show that $\left(  x_{n}\right)  $ is a Cauchy sequence, where
$$
x_{n}=\frac{\sin1}{2}+\frac{\sin2}{2^{2}}+\ldots+\frac{\sin n}{2^{n}}.
$$
We try to evaluate $\left\vert x_{n+p}-x_{n}\right\vert $ and to show that
this one is arbitrary small. So
\begin{align*}
\left\vert a_{n+p}-a_{n}\right\vert  & =\left\vert \frac{\sin\left(
n+1\right)  }{2^{n+1}}+\frac{\sin\left(  n+2\right)  }{2^{n+2}}+\ldots
+\frac{\sin\left(  n+p\right)  }{2^{n+p}}\right\vert \\
& \leq\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\ldots+\frac{1}{2^{n+p}}.
\end{align*}
Now, $\left(  \frac{1}{2^{n+j}}\right)  $ converge all to zero. Hence
$\left\vert a_{n+p}-a_{n}\right\vert \rightarrow0$. 
Does it work this argument according to the definition of Cauchy sequence 
$$
\forall\varepsilon>0,
\quad
\exists n_{\varepsilon}\in
\mathbb{N}
,
\quad
\forall n\in
\mathbb{N}
,
\quad
\forall p\in
\mathbb{N}
:n,p\geq n_{\varepsilon}\Rightarrow\left\vert a_{n+p}-a_{n}\right\vert
<\varepsilon?
$$
My question was born from the fact that for the sequence
$$
a_{n}=1+\frac{1}{2}+\ldots+\frac{1}{n},
$$
we have that
\begin{align*}
\left\vert a_{n+p}-a_{n}\right\vert  & =\left\vert \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+p}\right\vert \\
& \leq\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+p}
\end{align*}
and also all sequences $\left(  \frac{1}{n+j}\right)  _{n}$ converge to zero,
but this time $\left(  a_{n}\right)  $ is not a Cauchy sequence.

It is not sufficient that all $\left(  \frac{1}{2^{n+j}}\right)$ converge to zero. A correct argument would be  that
$$
\left\vert a_{n+p}-a_{n}\right\vert  \le
\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}+\ldots+\frac{1}{2^{n+p}}
 = \frac{1}{2^{n}} \left(  \frac 12 + \frac 14 + \ldots + \frac{1}{2^{p}}\right) < \frac{1}{2^{n}}
$$
becomes arbitrary small for large $n$ and arbitrary $p$.
That won't work for the harmonic series.
