Proving a sequence is a Cauchy sequence using definition I am intending to show that the sequence $<f_n>$ where $f_n$ = $ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}........+\frac{(-1)^{n-1}}{n}$ is a Cauchy sequence using the definition of a Cauchy sequence.
My attempt: Let $\epsilon>0$ and $n>m$ 
$|f_n-f_m|=|\frac{(-1)^m}{m+1}+\frac{(-1)^{m+1}}{m+2}........\frac{(-1)^{n-1}}{n}|$
I am stuck here for how do I find a particular m for each $\epsilon>0$ such that the conditions required for being a Cauchy sequence is satisfied.
 A: If $(n-m)$ is odd, we may assume $m$ is even without loss of generality, since $\left\lvert k \right\rvert = \left\lvert -k \right\rvert$. Then \begin{align*}\left\lvert f_n - f_m\right\rvert &= \left\lvert \frac{1}{m+1} - \frac{1}{m+2} + \frac{1}{m+3} - \frac{1}{m+4} + \dotsb + \frac{1}{n-2} - \frac{1}{n-1} + \frac{1}{n}\right\rvert \\ &= \frac{1}{m+1} - \frac{1}{m+2} + \dotsb + \frac{1}{n-2} - \frac{1}{n-1} + \frac{1}{n}\end{align*} since we can see that each pair of terms, of the form $\left(\frac{1}{k} - \frac{1}{k+1}\right)$, is positive, and the final term, $\frac{1}{n}$, is also positive, so the whole expression inside the absolute value signs is positive. Now, we can rewrite this as \begin{align*}\left\lvert f_n - f_m \right\rvert = \frac{1}{m+1} + \left(\frac{1}{m+3}-\frac{1}{m+2}\right) + \dotsb + \left(\frac{1}{n}-\frac{1}{n-1}\right)\end{align*} where each bracket is clearly negative (or zero), from which it follows that \begin{equation*}\left\lvert f_n - f_m \right\rvert \leq \frac{1}{m+1} \leq \frac{1}{m}\end{equation*}
In the case where $(n-m)$ is even, again assume $m$ is even without loss of generality. We can now re-use our working above, as we have \begin{align*}\left\lvert f_n - f_m \right\rvert &= \left\lvert f_{n} - f_{m+1} + f_{m+1} - f_{m}\right\rvert \\ &\leq \left\lvert f_{n} - f_{m+1}\right\rvert + \left\lvert f_{m+1}-f_{m}\right\rvert \qquad \text{(by the triangle inequality)} \\ &= \left\lvert f_{n} - f_{m+1}\right\rvert + \frac{1}{m+1} \\ &\leq \frac{1}{m+1} + \frac{1}{m+1} \qquad \text{(by the above working, as $(n-(m+1))$ is odd)} \\ &\leq \frac{2}{m}\end{align*}
Since $\frac{1}{m} \leq \frac{2}{m}$, it follows that you can make $\left\lvert f_n - f_m \right\rvert$ less than any desired $\epsilon > 0$ by taking $n > m > \frac{2}{\epsilon}$, so you are done.
