# showing $\sum_{n=1}^\infty (y_n-y_{n+1})$ converges $\Leftrightarrow$ $(y_n)_{n\in\mathbb N}$ converge

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?

• There is no need for an Cauchy sequence argument. Just calculate partial sum of $x_n$. – Du Phan Dec 16 '13 at 17:21
• This has nothing to do with complex analysis. – mrf Dec 16 '13 at 19:51

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.
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 $$\sum_{n=1}^\infty(y_n-y_{n+1})=\lim_nS_n=y_1-y_\infty.$$
• 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. – Andrés E. Caicedo Dec 16 '13 at 18:59