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Prove that the following succession is Cauchy {$x_{n}={\frac{cos1}{3}+\frac{cos2}{3^{2}}+...+\frac{cosn}{3^{n}}}$}

Only I have to show that this sequence converges and a theorem already proved "any convergent sequence is Cauchy"

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Can you use results on infinite series? In particular, on convergent geometric series and the comparison test? –  David Mitra Apr 24 '12 at 1:22
    
Did you care about the answers you got in previous questions ? It seems like you didn't... –  Lierre Apr 24 '12 at 1:30
    
If I can use the infinite series –  Miguel Mora Luna Apr 24 '12 at 1:33

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up vote 2 down vote accepted

You can show the sequence is Cauchy directly: $$\eqalign{|x_{m+n}-x_n|&= \Biggl| { {\cos (n+1)\over 3^{n+1}} + { {\cos (n+2)\over 3^{n+2}} } +\cdots+{\cos(m+n)\over 3^{m+n}} } \Biggr|\cr &\le \Biggl| { {\cos (n+1)\over 3^{n+1}} \Biggr|+ \Biggl| { {\cos (n+2)\over 3^{n+2}} }\Biggr| +\cdots+\Biggl|{\cos(m+n)\over 3^{m+n}} } \Biggr|\cr &\le { {1\over 3^{n+1}} + {1\over 3^{n+2} } +\cdots+ {1\over 3^{m+n}} } \cr &={ {1\over3^{n+1}}-{1\over 3^{m+n+1}} \over 1-(1/3) }\cr &={ {1\over2\cdot 3^{n}}-{1\over 2\cdot 3^{m+n}} }. }$$


If you want to show that $(x_n)$ converges, note that $x_m$ is the $m$'th partial sum of $\sum\limits_{n=1}^\infty {\cos n \over 3^n}$. So the sequence $(x_n)$ converges if and only if the sum $\sum\limits_{n=1}^\infty {\cos n \over 3^n}$ converges.

If you can use standard results concerning infinite series, then you can show $\sum\limits_{n=1}^\infty { \cos n \over 3^n}$ converges absolutely, and thus converges, by applying the Comparison Test to the series $\sum\limits_{n=1}^\infty {|\cos n|\over 3^n}$ and the convergent geometric series $\sum\limits_{n=1}^\infty {1\over 3^n}$. Note here that ${|\cos m|\over 3^m}\le {1\over 3^m}$.

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thank you very much –  Miguel Mora Luna Apr 24 '12 at 1:44

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