Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n \in \mathbb{N}$, and $(x_n) \to 0$. Let $(y_n)$ be defined for all $n \in \mathbb{N}$ by
$$y_n = x_0 – x_1 + x_2 – \cdots + (-1)^n x_n \ .$$
I want to show, using the $\varepsilon$ definition, that $(y_n)$ is Cauchy.
I am trying to find, given $\varepsilon > 0$, a real number $N$ such that for all $m$ and $n$ with $m > n > N$, $|y_m – y_n| < \varepsilon$.
I have been going backwards to try and find $N$, and have
\begin{align*}
|y_m – y_n| & = \left| (x_0 – x_1 + \cdots \pm x_m) – (x_0 – x_1 + \cdots \pm x_n) \right| \\
|y_m – y_n| & = \left| x_{n + 1} – x_{n + 2} + \cdots \pm x_{m} \right| \\
|y_m – y_n| & \leq | x_{n + 1} | + | x_{n + 2} | + \cdots + | x_{m} | \\
|y_m – y_n| & \leq \ ?
\end{align*}
I do not know how to get a solution from there, and am not sure about the process, particurlary the last step since I feel getting rid of the minuses might prevent me from finding a solution.
Best Answer
To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $\sum_{k=1}^{\infty}\frac{(-1)^k}{k}$.
What you can do is grouping the terms of the partial sums $s_n= \sum_{j=1}^n(-1)^jx_j$ as follows:
Now, you can write $|s_{m} - s_n|$ in two different ways:
$$|s_{n+k} - s_n| = \begin{cases} |x_{n+1} - (x_{n+2}-x_{n+3}) - \cdots - (x_{n+2i}-x_{n+2i+1})| & k = 2i+1 \\ |x_{n+1} - (x_{n+2}-x_{n+3}) - \cdots - (x_{n+2i-2}-x_{n+2i-1}) - x_{2i}| & k = 2i \\ \end{cases} $$
$$|s_{n+k} - s_n| = \begin{cases} |(x_{n+1} - x_{n+2}) + \cdots + (x_{n+2i-1}-x_{n+2i}) + x_{n+2i+1}| & k = 2i+1 \\ |(x_{n+1} - x_{n+2}) + \cdots + (x_{n+2i-1}-x_{n+2i}) | & k = 2i \\ \end{cases} $$
Using the fact that $x_n \searrow 0$, it follows immediately that for all $k \in \mathbb{N}$ holds $$|s_{n+k} - s_n| \leq x_{n+1}$$
Hence, for $\epsilon > 0$ choose $N_{\epsilon}$ such that $x_{N_{\epsilon}} < \epsilon$. Then, for all $m> n > N_{\epsilon}$ you have $$|s_{m} - s_n| \leq x_{n+1} \leq x_{N_{\epsilon}} < \epsilon$$