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Let $\{x_n\}, n\in\Bbb N$ denote a sequence such that the sequence: $$ \left\{\sum_{k=1}^n x_n \right\} $$ converges. Prove that: $$ \lim_{n\to\infty}x_n = 0 $$

Please note that it is $x_n$ under the sum sign, which i believe is a typo, isn't it? I have replaced it with $x_k$ below.

Let: $$ y_n = \left\{\sum_{k=1}^nx_k \right\} $$

We know that $y_n$ converges, hence is satisfies the Cauchy criteria: $$ \forall \epsilon > 0\ \exists N\in\Bbb N: \forall n,m > N \implies |y_n - y_m| < \epsilon $$

Consider $|y_n - y_m|$ for $m>n$: $$ \begin{align} |y_n - y_m| &= |y_m - y_n| \\ &= \left|\sum_{k=n+1}^m x_k\right| \\ &\ge \sum_{k=n+1}^m |x_k| \\ &= |x_{n+1}| + |x_{n+2}| + \cdots + |x_m| \\ &\ge |x_{n+1}| \end{align} $$

So we have: $$ |x_{n+1}| \le |y_n - y_m| < \epsilon $$

Which means: $$ \forall \epsilon > 0\ \exists N\in\Bbb N: \forall n> N \implies |x_{n+1}| < \epsilon $$

But that is the definition of a limit, thus: $$ \lim_{n\to\infty} x_n = 0 $$

Is my proof valid? Also is it really a typo in the book or am I missing something?

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  • $\begingroup$ A bit easier: $x_n = y_{n+1} - y_n$, and since the series $(y_n)_n$ converges, the limit is $\lim \limits_{n \to \infty} x_n = \lim \limits_{n \to \infty} (y_{n+1} - y_n) = 0$. $\endgroup$ – ComplexFlo Dec 24 '18 at 14:09
  • $\begingroup$ @ComplexFlo I think you mean $x_{n+1}$ but of course the same argument applies with the Shift Rule. $\endgroup$ – AlephNull Dec 24 '18 at 14:10
  • $\begingroup$ @AlephNull you're completely right, it should be $x_{n+1}$ actually! But the argument stays valid anyway. Thanks for the hint! $\endgroup$ – ComplexFlo Dec 25 '18 at 9:54
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The proof is not valid since your triangle inequality is reversed in sign.

That is $$\left| \sum_{k=n+1}^mx_k \right| \le\sum_{k=n+1}^m|x_k| $$

You might like to use Cauchy condition on $y_n$ and show that

$$|y_n - y_{n-1}|=|x_n|$$

can be arbitrarily close to $0$.

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