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I want to show for $x_n:=y_n-y_{n+1}\in\mathbb C$:

$\sum_{n=1}^\infty x_n$ converges $\Leftrightarrow$ $(y_n)_{n\in\mathbb N}$ converge

I've tried the following:

$(y_n)_{n\in\mathbb N}$ converge $\Leftrightarrow$ $(y_n)_{n\in\mathbb N}$ is a Cauchy-sequence $\Leftrightarrow $ for all $\epsilon>0$ there is a $n_0\in\mathbb N$ such that for alle $n\geq m\geq n_0$: $|y_m-y_{n+1}|<\epsilon$ $\Leftrightarrow$ $\forall\epsilon>0\exists n_0\in\mathbb N:\forall m\geq n\geq n_0:\left|\sum_{k=m}^nx_k\right|<\epsilon$ $\Leftrightarrow$ $\sum_{n=1}^\infty x_n$ converges by Cauchy

Can you do it this way? And are there any easier ways if my one is incorrect?

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    $\begingroup$ There is no need for an Cauchy sequence argument. Just calculate partial sum of $x_n$. $\endgroup$ – Du Phan Dec 16 '13 at 17:21
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    $\begingroup$ This has nothing to do with complex analysis. $\endgroup$ – mrf Dec 16 '13 at 19:51
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The series $\displaystyle\sum_{n=1}^\infty y_n-y_{n+1}$ is convergent if and only if the partial sum $\displaystyle \sum_{k=1}^{n-1} y_k-y_{k+1}=y_1-y_n$ is convergent which means that $(y_n)_n$ is convergent.

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  • $\begingroup$ thanks, what about my attempt? $\endgroup$ – user104143 Dec 16 '13 at 17:33
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    $\begingroup$ As said in Du Phan's comment there's no need for an Cauchy sequence argument. $\endgroup$ – user63181 Dec 16 '13 at 17:35
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    $\begingroup$ Wow, you've had a busy day or two! +1 $\endgroup$ – Namaste Dec 17 '13 at 17:54
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Let $$ S_n=\sum_{i=1}^n(y_i-y_{i+1})=\sum_{i=1}^ny_i-\sum_{j=2}^{n+1}S_j=y_1-y_{n+1}. $$ If $\sum_{n=0}^\infty(y_n-y_{n+1})$ converges, i.e. $$ S:=\lim_nS_n=\sum_{n=0}^\infty(y_n-y_{n+1})<\infty, $$ then we have $$ \lim_ny_n=\lim_n(y_1-S_{n-1})=y_1-S. $$ Conversely, if $(y_n)$ converges, say with $y_\infty=\lim_ny_n$, then $$ \sum_{n=1}^\infty(y_n-y_{n+1})=\lim_nS_n=y_1-y_\infty. $$

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    $\begingroup$ Do not use $\sum_n(y_n-y_{n-1})<\infty$ as synonym for "exists". This is only correct if the terms of the series are positive. Here, they are complex numbers. $\endgroup$ – Andrés E. Caicedo Dec 16 '13 at 18:59

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