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

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.

• Because the series is alternating and decreasing, I think you can prove by induction on $m$ that $|y_m-y_n| \leq |y_n|$. – Robert Shore Apr 25 at 1:17
• @RobertShore is my answer okay? – Subhasis Biswas Apr 25 at 1:37
• @RobertShore yes I can definitely show that, but it brings me to the same issue with $|y_m| \leq |x_0 - x_1 + \cdots \pm x_m|$, and I am unsure how to proceed from there. – oranji Apr 25 at 4:05
• I meant to say you can prove by induction that $|y_m-y_n| \leq |x_n|$. Since $\lim x_n=0$, choose $N$ such that $n \gt N \Rightarrow |x_n| \lt \epsilon$. Then $|y_m-y_n| \leq |x_n| \lt \epsilon$ so $\{y_n\}$ is Cauchy. – Robert Shore Apr 25 at 5:54

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

• This is exactly what I was about to do. – Subhasis Biswas Apr 25 at 4:06
• @SubhasisBiswas So, I did it for you :-D – trancelocation Apr 25 at 4:07

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.

• I want to use the $\varepsilon$ definition of a Cauchy sequence, and not the fact that all convergent sequences are Cauchy, which is why I cannot use this solution. – oranji Apr 25 at 4:02
• I'll edit this answer. – Subhasis Biswas Apr 25 at 4:05