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.
