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Let $a_n$ be any convergent sequence of real numbers and let $x_n = a_{n+1} − a_n$ for each $n \in\mathbb{N}$. Prove that the sum $x_n$ as $n$ goes to infinity is a convergent series and find its sum.

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@Norbert: I am not sure whether it is good to use \limits in the title - for the reasons similar to: Why no use displaystyle in titles?. –  Martin Sleziak Jul 11 '12 at 17:41
@MartinSleziak, \limits often used in question om MSE, and for my taste I don't like \sum signs without specification range of summing index. If you don't agree you can rollback or edit it in an appropriate way –  userNaN Jul 11 '12 at 17:49
@Norbert You're right, using \limits in the titles seems to be ok. (I remembered this was discussed somewhere and found this.) –  Martin Sleziak Jul 11 '12 at 18:56

1 Answer 1

This is called a telescopic sum. Try to write explicit expression for the $\sum\limits_{k=1}^N x_k$. You will see that most terms will cancel out. Then take a limit when $N\to \infty$.


$$ \sum\limits_{k=1}^N x_k = (a_2-a_1)+(a_3-a_2)+(a_4-a_3)+\ldots+(a_{N-2}-a_{N-3})+(a_{N-1}-a_{N-2})= $$ $$ a_2-a_1+a_3-a_2+a_4-a_3+\ldots+a_{N-2}-a_{N-3}+a_{N-1}-a_{N-2}=a_{N}-a_1 $$

We have just computed the partial sum. In order to pass to the infinite series take limit as $N \rightarrow \infty$. $$ \sum\limits_{k=1}^\infty x_k= \lim\limits_{N\to\infty}\left(\sum\limits_{k=1}^N x_k\right)= \lim\limits_{N\to\infty}\left(a_N-a_1\right)= \left(\lim\limits_{N\to\infty}a_N\right)-a_1 $$

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could you elaborate more please? –  max Feb 27 '12 at 16:47

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