May 1999 #1

Problem Statement: Let \{a_n\} be a sequence of real numbers. Prove that \{a_n\} converges if and only if \sum\limits_{n=0}^{\infty}(a_{n+1}-a_n) converges.

Proof: (\Rightarrow) Assume that \{a_n\} converges to some real number L. Let \varepsilon>0. Then there exists an N\in\mathbb{N} such that |a_n-L|<\varepsilon for every n\geq N.

Consider \{S_N=\sum\limits_{n=0}^{N}\}(a_{n+1}-a_n)\}S_N=(a_1-a_0)+(a_2-a_1)+\dots+(a_{N+1}-a_N)=-a_0+a_{N+1}.

Let n\geq N, then |S_n-(L-a_0)|=|-a_0+a_{n+1}-L+a_0|=|a_{n+1}-L|<\varepsilon since n+1>N. Thus, \{S_n\} converges and so the series \sum\limits_{n=0}^{\infty}(a_{n+1}-a_n) converges.

(\Leftarrow) Assume that \sum\limits_{n=0}^{\infty}(a_{n+1}-a_n) converges to M.

Let \varepsilon>0. Then there exists some K\in\mathbb{N} such that |S_k-M|<\varepsilon for every k\geq K. As shown above S_k=-a_0+a_{k+1}, so, for k\geq K it follows that |-a_0+a_{k+1}-M|<\varepsilon. This means that \{a_k\} converges to M+a_0 since a_0 is a fixed real number.


Reflection: This one really fell straight out from using the definitions. I’m starting to learn that it’s often a good idea to just try to do a proof directly from the definition. What’s really happening here is that the terms are getting really close together. Since they’re close together their difference is going to zero, hence the series converges. But, if you know that the difference is going to zero that means that the terms in the sequence are getting close, so the sequence converges.

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