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Suppose $|a_n| \leq b_n - b_{n+1}$ where $b_n$ decreases monotonically to zero. Prove that $\sum_{n=1}^{\infty} a_n$ converges absolutely.

My thoughts were $\sum_{n=1}^{\infty} |a_n| \leq \sum_{n=1}^{\infty} (b_n - b_{n+1})$.

Now I'd to use the fact $b_n$ is monotonically decreasing to zero to show the sum on the right converges so by comparison so must the sum on the left, which proves what I want. My trouble is in formally proving that the right sum converges.

Informally I know $b_n \geq b_{n+1} \geq b_{n+2} ...$ and eventually for some $k$, $b_k = 0$, so we just have a finite number of terms we're summing up so it converges to whatever that is. I don't know if I can take for granted that the sequence hits $0$ at a finite $k$. Even then the details seem a bit hazy.

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Remind the definition of infinite sum: $$\sum_{k=1}^\infty (b_k-b_{k+1})=\lim_{n\to\infty}\sum_{k=1}^n (b_k-b_{k+1})=\lim_{n\to\infty}b_1-b_{n+1}.$$

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  • $\begingroup$ Oh. I see the idea is if you go out to $n$ every other term cancels out, so you're left with the limit you've written and for $n$ large enough $b_{n+1}$ = 0, so we have the sum equals $b_1$. Thanks! $\endgroup$ – Eddie Nov 7 '14 at 15:53

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