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?
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 $0$.