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## Proof: (related to "Convergence Test for Telescoping Series")

• The telescoping partial sum equals $$\begin{array}{rcl}\sum_{k=0}^n (b_k-b_{k+1})&=&(b_0-b_1)+(b_1-b_2)+(b_2-b_3)+\ldots—b_{n}+(b_n-b_{n+1})\\&=&b_0+(-b_1+b_1)+(-b_2+b_2)+-b_3+\ldots+(-b_{n}+b_n)-b_{n+1}\\&=&b_0-b_{n+1}.\end{array}$$
• Since, by hypothesis, $(b_k)_{k\in\mathbb N}$ is convergent sequence, the infinite series $\sum_{k=0}^\infty (b_k-b_{k+1})$ converges to the value $b_0-\beta,$ where $\beta:=\lim_{k\to\infty} b_k.$
• If, in addition, $(b_k)_{k\in\mathbb N}$ is monotonic, the terms $(b_k-b_{k+1})$ are either all $\ge 0$ or all $\le 0.$
• In the case $(b_k-b_{k+1})\ge 0$ we have that $\sum_{k=0}^\infty |b_k-b_{k+1}|$ is absolutely convergent.
• In the case $(b_k-b_{k+1})\le 0$ we have $|b_k-b_{k+1}|=|b_{k+1}-b_k|=b_{k+1}-b_k.$
• Therefore, $\sum_{k=0}^\infty |b_k-b_{k+1}|$ is also absolutely convergent, since the infinite series $\sum_{k=0}^\infty (b_{k+1}-b_k)$ converges to $\beta-b_0.$
q.e.d

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