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Proposition Suppose $\{a_n\}$ and $\{b_n\}$ are real sequences such that $a_n \to 0$. Show that $\sum a_kb_k$ converges under the following two conditions: $\sum_{k=1}^{\infty} |a_{k+1}-a_k|< \infty$ and $|\sum_{k=1}^{n} b_k| \leq M$.

I'm stuck on this proof because I'm not sure how this: $\sum_{k=1}^{\infty} |a_{k+1}-a_k|< \infty$ fits into the bigger picture. In terms of finite sums this would be a telescoping series.

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Suppose that $\sum_k|a_{k+1}-a_k|<\infty$ and $|\sum_{k=1}^nb_k|\leq M$ (I will assume that this last condition happens for all $n$).

Using Newton series, for $n>m$, $$ \sum_{k=m}^na_kb_k=a_n\sum_{k=m}^nb_k-\sum_{k=m}^{n-1}(a_{k+1}-a_k)\sum_{h=m}^kb_h, $$ so, $$ \left|\sum_{k=m}^na_kb_k\right|\leq\,|a_n|\,\left|\sum_{k=m}^nb_k\right|+\sum_{k=m}^{n-1}|a_{k+1}-a_k|\,\left|\sum_{h=m}^kb_h\right|\leq\,M\,|a_n|+M\,\sum_{k=m}^{n-1}|a_{k+1}-a_k|. $$ Given $\varepsilon>0$, we can choose $m$ such that $|a_n|<\varepsilon/2M$, $\sum_{k=m}^n|a_{k+1}-a_k|<\varepsilon/2M$, which yields $$ \left|\sum_{k=m}^na_kb_k\right|<\varepsilon. $$ So the series converges.

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Hint: Let $S_n= \sum_{k=1}^{n}b_n$ and construct the partial sum

$$ V_n = \sum_{k=0}^{n}a_n b_n = a_1 b_1 + a_2 b_2 + \dots + a_n b_n $$

$$ = a_1 S_1 + a_2 (S_2-S_1) + \dots + a_n (S_n-S_{n-1}) $$

$$ = S_1(a_1-a_2)+ S_2(a_2-a_3)+ \dots + S_{n-1}(a_{n-1}-a_n)+ a_n S_n $$

$$ \implies V_n= \sum _{k=1}^{n-1}S_k (a_k-a_{k-1}) + a_n S_n .$$

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