# Proof verification that if $y_n = \sum_{k=1}^n x_n$ converges then $\lim_{n\to\infty} x_n = 0$

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?

• A bit easier: $x_n = y_{n+1} - y_n$, and since the series $(y_n)_n$ converges, the limit is $\lim \limits_{n \to \infty} x_n = \lim \limits_{n \to \infty} (y_{n+1} - y_n) = 0$. – ComplexFlo Dec 24 '18 at 14:09
• @ComplexFlo I think you mean $x_{n+1}$ but of course the same argument applies with the Shift Rule. – AlephNull Dec 24 '18 at 14:10
• @AlephNull you're completely right, it should be $x_{n+1}$ actually! But the argument stays valid anyway. Thanks for the hint! – ComplexFlo Dec 25 '18 at 9:54

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 $$0$$.