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Let $a_n \rightarrow 0$. Show $\sum\limits_{n=0}^\infty (a_n - a_{n+1}) = a_0$.

In the case that $\sum\limits_{n=0}^\infty a_n$ is convergent, I was able to show that $$\sum\limits_{n=0}^\infty (a_{n} - a_{n+1}) = \sum\limits_{n=0}^\infty a_n - \sum\limits_{n=0}^\infty a_{n+1} = a_0 + \sum\limits_{n=1}^\infty a_n - \sum\limits_{n=0}^\infty a_{n+1} = a_0 + \sum\limits_{n=0}^\infty a_{n+1} - \sum\limits_{n=0}^\infty a_{n+1} = a_0. $$

For the case where $\sum\limits_{n=0}^\infty a_n$ is divergent, I am not really sure what to do.

I was thinking that since $a_n \rightarrow 0$, we have that for all $\epsilon > 0$, there exists $k \in \mathbb{N}$ such that $|a_n| < \epsilon$ for all $n \geq k$.

So, then we get that $$\sum\limits_{n=0}^\infty (a_n - a_{n+1}) = a_0 - a_k + \sum\limits_{n=k}^\infty (a_n - a_{n+1}).$$ This converges if and only if the series of partial sums converges, and we have that $$S_k = a_0 - a_k + \sum\limits_{n=k}^K (a_n - a_{n+1}).$$ So, taking the limit as $k$ goes to infinity, we get $a_0$? Why would the last term go away?

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Note that $$\sum_{n=0}^N(a_n-a_{n+1})=a_0-a_{N+1}, \forall N\in \mathbb{N}.$$ So if $a_n\to 0$ we can conclude that the series is convergent and its sum is $a_0.$

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  • $\begingroup$ Oh, duh. Thanks! $\endgroup$ – user389056 Nov 14 '16 at 1:19
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This is a telescoping series. $S_k=a_0-a_{k+1}$. Then take limit for your answer.

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