Prove the alternating sum of a decreasing sequence converging to $0$ is Cauchy.

Question

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.

Answer 1

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:

• Let $m = n+k, k,n \in \mathbb{N}$

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 $$|s_{m} - s_n| \leq x_{n+1} \leq x_{N_{\epsilon}} < \epsilon$$

Answer 2

This is also known as the "Leibnitz's Test".

We write $s_n = x_1-x_2+x_3-...+(-1)^{n+1}x_n$

$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2} \geq0$ for all $n$.

$s_{2n+1}-s_{2n-1}=-u_{2n}+u_{2n+1} \leq 0$

$s_{2n} =u_1 -(u_2-u_3)-(u_4-u_5)...-u_{2n} \leq u_1$, i.e. a monotone increasing sequence bounded above.

$s_{2n+1} =(u_1 -u_2)+(u_3-u_4)+...+u_{2n+1} \geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.

Hence, both are convergent subsequences of $(s_n)$. But, we have $\lim (s_{2n+1}-s_{2n})=u_{2n+1}=0$, therefore, they converge to the same limit.

Hence, $(s_n)$ converges, i.e. it is Cauchy.

Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_{2n})$ and $(s_{2n+1})$ i.e. $U =\{ 2n+1 : n \in \mathbb{N}\}$ and $V =\{ 2n : n \in \mathbb{N}\}$ form a partition of $\mathbb{N}$ and they both converge to the same limit.