Let $\{x_n\}$ be a sequence of real numbers. Suppose that for each $\epsilon > 0$ there is an $N \in \mathbb{N}$ such that $m \geq n \geq N$ implies $ |\sum_{k=n} ^m x_k | < \epsilon$.
Prove that $$ \lim_{n\to \infty} \sum_{k=1}^n x_k $$ exists and is finite.
Intuitively I believe I want to be able to show $$ \lim_{n\to \infty} \sum_{k=1}^n x_k = \sum_{k=1}^{N-1} x_k + \lim_{n\to \infty} \sum_{k=N}^n x_k$$
The first sum is finite. It's trying to show that the second limit is 0. I've used the case where $n=m$ to show $|x_n|<\epsilon$. Similarly $|x_n+x_{n+1}|<\epsilon$, and $|x_n+x_{n+1}+...x_m|<\epsilon$. Am I wrong in thinking that I've shown $x_n \to \infty$ as $n \to \infty$. We also know that the sum of any terms after N are arbitrarily close. I'm not certain how I can show that the sum implied in the beginning is equal to 0.
I like, on questions like these, to go back to the definition, in this case of "a limit exists, and is finite."
For a sequence $s_n$, a limit exists, and is finite, if there is some value $s$ such that for any arbitrary real $\varepsilon > 0$ there exists some integer $K$ (which is allowed to depend on the selected $\varepsilon$) such that $k > K \implies |s-s_k| < \varepsilon$.
So our task is to find some (function) $K(\varepsilon)$ that meets this definition for our sequence $x_k$.
From the given, for any $\epsilon$, there exists some $N(\epsilon)$ such that $m >= n >= N(\epsilon) \implies |\sum_n^m x_k | < \epsilon$. Since all allowed values of $N(\epsilon)$ are positive we can further refine that definition to say that $N(\epsilon$ is the least such integer. In this statement, we can choose, for example, various values of $\epsilon$: $$ \epsilon_1 = \varepsilon/2 \\ \epsilon_2 = \varepsilon/2^2 \\\cdots $$ and we would then have a sequence $N(\epsilon_1), N(\epsilon_2), \cdots$.
Now choose the $n$ and $m$ in each $|\sum_n^m x_k |$ to be $N(\epsilon_i)$ and $N(\epsilon_{i_1})$, respectively. Then $$ \sum_1^L x_k = \sum_1^{N(\epsilon_1)} x_k + \left( \sum_{N(\epsilon_1)}^{N(\epsilon_2)} x_k + \sum_{N(\epsilon_2)}^{N(\epsilon_3)} x_k + \cdots \right) $$ Now consider the expression in the large parentheses: In absolute value, the terms are less than $$ \epsilon_1, \epsilon_2, \ldots = \varepsilon/2, \varepsilon/2^2, \varepsilon/2^3 \ldots $$ so the absolute value of the sum is less than $\varepsilon\left(\frac12+\frac14+\frac18\cdots\right) = \varepsilon$.
So we have shown that be choosing $K(\varepsilon)$ in our definition to be $N(\varepsilon/2)$ in the given, the series $x_i$ meets the definition and thus has a finite limit.