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Let $x_1, x_2, \ldots$ be a Cauchy sequence in a normed vector space $X$. Suppose that any absolutely convergent series converges. Then: prove that $x_1, x_2, \ldots$ converges.

My idea was: set $y_n = x_n - x_{n-1}$, then if we can prove that the sum of the $y_n$ converges absolutely, this means that $$ \sum_{n = 1}^{\infty} y_n = \lim_{k \to \infty} \sum_{n = 1}^{k} y_n = \lim_{k \to \infty} x_k $$ exists. But the problem is: how do I prove that $$ \sum_{n = 1}^{\infty} ||y_n|| $$ converges? Or is this the wrong approach? (In the latter case: don't spoil too much, please.)

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  • $\begingroup$ In general, your series of $y_n$ will not be absolutely convergent. But (using the fact that $x_n$ is Cauchy) you need to do something like this, but first extract a good subsequence of $x_n$. $\endgroup$
    – GEdgar
    Oct 6, 2016 at 13:28

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Hint : It's actually a very good approach. But, there's actually no guarantee that $\sum_n ||x_n - x_{n+1}||$ is converging. Nonetheless, because $(x_n)_n$ is a Cauchy sequence, you can extract a subsequence $(x_{n_k})_k$ such that $||x_{n_k} - x_{n_{k+1}}|| \leq 2^{-k}$ (Exercise 1).

Then $\sum_k ||x_{n_k} - x_{n_{k+1}}|| \leq \sum_k 2^{-k} < +\infty$

Therefore $(x_{n_k})_k$ is converging. Now, a Cauchy sequence that has a converging subsequence is convergent (Exercise 2), thus leading to the conclusion.

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