Proving a series is Cauchy

Question

I would like to show that the sequence of partial sums of a series is Cauchy. The series is as follows: $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}$$ I am not allowed to use the fact that a series being convergent implies that it is Cauchy. I am trying to use the Cauchy Criterion for Series instead. I know that because of this I will work from: $$\left\lvert\sum_{k=m+1}^{n}\frac{(-1)^{k+1}}{k}\right\rvert$$ and I will have to show that this is less than all $\epsilon > 0$. I'm stuck at where to go from here. I'd like to find a series that is greater than or equal to the one given and then show that that is less than $\epsilon$, but I can't think of a series that is similar. I think that doing $\frac{1}{k}$ would be too dissimilar to the original series. Would I need to define a specific $N$ for when $n>m\ge N$? I also thought about making the series into $\lvert s_{m+1}+s_{m+2}+\cdots+s_n \rvert$ and then breaking it up by the triangle inequality, but after that I don't know where I'd go either. I would appreciate it if anyone could point me in the right direction.

Answer 1

Suppose that $m$ is even. Then$$\sum_{k=m+1}^n\frac{(-1)^{k+1}}k=\frac1{m+1}-\frac1{m+2}+\cdots+\frac{(-1)^{n+1}}n.\tag1$$But\begin{align}(1)&=\frac1{m+1}-\left(\frac1{m+2}-\frac1{m+3}\right)-\left(\frac1{m+4}-\frac1{m+5}\right)-\cdots\\&<\frac1{m+1}\end{align}and\begin{align}(1)&=\left(\frac1{m+1}-\frac1{m+2}\right)+\left(\frac1{m+3}-\frac1{m+4}\right)+\cdots\\&>\frac1{m+1}-\frac1{m+2}\\&>0.\end{align}So,$$\left|\sum_{k=m+1}^n\frac{(-1)^{k+1}}k\right|<\frac1{m+1}$$and the same thing occurs if $m$ is odd. So, given $\varepsilon>0$, choose $N$ such that $\frac1{N+1}<\varepsilon$.