Proof
(related to Lemma: Convergence Test for Telescoping Series)
 The telescoping partial sum equals $$\begin{array}{rcl}\sum_{k=0}^n (b_kb_{k+1})&=&(b_0b_1)+(b_1b_2)+(b_2b_3)+\ldotsb_{n}+(b_nb_{n+1})\\&=&b_0+(b_1+b_1)+(b_2+b_2)+b_3+\ldots+(b_{n}+b_n)b_{n+1}\\&=&b_0b_{n+1}.\end{array}$$
 Since, by hypothesis, $(b_k)_{k\in\mathbb N}$ is convergent sequence, the infinite series $\sum_{k=0}^\infty (b_kb_{k+1})$ 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 $(b_kb_{k+1})$ are either all $\ge 0$ or all $\le 0.$
 In the case $(b_kb_{k+1})\ge 0$ we have that $\sum_{k=0}^\infty b_kb_{k+1}$ is absolutely convergent.
 In the case $(b_kb_{k+1})\le 0$ we have $b_kb_{k+1}=b_{k+1}b_k=b_{k+1}b_k.$
 Therefore, $\sum_{k=0}^\infty b_kb_{k+1}$ is also absolutely convergent, since the infinite series $\sum_{k=0}^\infty (b_{k+1}b_k)$ converges to $\betab_0.$
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References
Bibliography
 Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition