Proof that a sequence is Cauchy.
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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 $leftvert x_{n+p}-x_{n}rightvert $ and to show that this one is arbitrary small. So begin{align*} leftvert a_{n+p}-a_{n}rightvert & =leftvert frac{sinleft( n+1right) }{2^{n+1}}+frac{sinleft( n+2right) }{2^{n+2}}+ldots +frac{sinleft( n+pright) }{2^{n+p}}rightvert \ & leqfrac{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 $leftvert a_{n+p}-a_{n}rightvert rightarrow0$ . Does it work this argument according to the definition of Cauchy sequence $$ forallvarepsilon>0, quad exists n_{varepsilon}in mathbb{N} , quad forall nin mathbb{N} , quad foral...