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Let $(a_n)_{n \in \mathbb{N}}$ and $(s_n)_{n \in \mathbb{N}}$ be sequences in $[0, \infty)$ so that $\sum\limits_{n=0}^{\infty}s_k$ converges and

$a_n \geq a_{n+1} -s_n ~~~~~~(n \in \mathbb{N})$

Show that $(a_n)_{n \in \mathbb{N}}$ is convergent.

A hint is given, i should sight the sequence $(b_n)_{n \in \mathbb{N}}$ with $b_n = a_n-\sum\limits_{n=0}^{n-1}s_k$ for $n \in \mathbb{N}$ first.

My idea is to show that the sequence is bounded above and monotonically nonincreasing. But how to do this?

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  • $\begingroup$ Consider that $b_n$ is nonincreasing $\endgroup$ – Farshad Nahangi Dec 27 '13 at 21:57
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We have $$b_{n+1}-b_n=a_{n+1}-a_n-s_n\le0$$ so the sequence $(b_n)$ is decreasing and $$b_n=a_n-\sum_{n=0}^{n-1}s_k\ge a_n-\sum_{n=0}^\infty s_k\ge-\sum_{n=0}^\infty s_k=C$$ hence $(b_n)$ is bounded below by $C$ so it's convergent. The convergence of $(a_n)$ can be deduced easily.

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  • $\begingroup$ You have the nice skill of saying "enough" without saying "too much"! +1 $\endgroup$ – amWhy Dec 28 '13 at 15:52

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