Let $x_n\to x$ and $k\in \mathbb N$. Consider the infinite series $\sum_{n=0}^\infty a_n$ where $a_n=x_n-x_{n+k}$. Then the partial sum $s_{mk-1}$ is equal to
$$s_{mk-1}=x_0+x_1+...+x_{k-1}-(x_{mk}+x_{mk+1}+...+x_{(m+1)k-1})$$
Thus
$$\sum_{n=0}^\infty a_n=\lim_{m\to\infty}s_{mk-1}=x_0+x_1+...+x_{k-1}-kx.$$
Now suppose that $c_1,...,c_k$ be $k\ge2$ numbers such that $c_1+...+c_k=0$ and $a_n=c_1x_{n+1}+c_2x_{n+2}+...+c_kx_{n+k}$. I'm trying to find the suitable partial sum to prove that
$$\sum_{n=0}^\infty a_n=c_1x_1+(c_1+c_2)x_2+...+(c_1+c_2+...+c_{k-1})x_{k-1}+(c_2+2c_3+3c_4+...+(k-1)c_k)x$$
Could anyone help me to find the suitable partial sum? What other methods can we use here? Thanks!
I think the best way to visuallize is to arrange all the terms in a grid: $$ \begin{align} a_1+\dots+a_n=\quad &c_1x_1+c_2x_2+c_3x_3+\dots+c_kx_k \\+&c_1x_2+c_2x_3+c_3x_4+\dots+c_kx_{k+1} \\+&c_1x_3+c_2x_4+c_3x_5+\dots+c_kx_{k+2} \\&\vdots \\+&c_1x_n+c_2x_{n+1}+c_3x_{n+2}+\dots+c_kx_{n+k-1} \end{align} $$ To exavulate this sum, collect all terms coerresponding to the same $x_i$, for each $i$. These are grouped in upward sloping diagonals in the above grid. Recalling that $\sum c_i=0$, $$ \begin{align} a_1+\dots+a_n &=c_1x_1+(c_1+c_2)x_2+\dots+(c_1+\dots+c_{k-1})x_{k-1} \\&\,\,\,\,+0x_k+0x_{k+1}+\dots+0x_n+ \\&\;\;+(c_2+\dots+c_k)x_{n+1}+\dots+(c_{k-1}+c_k)x_{n_k-1}+c_kx_{n+k-1} \end{align} $$ This is the formula for the partial sums. Letting $n\to\infty$ gives the formula for the infinite sum.