Show that $\langle f_n \rangle$ is a Cauchy sequence, where $f_n=1-\frac12+\frac13-\frac14+\dots+\frac{(-1)^{n-1}}{n}$ Show that $\langle f_n \rangle$, where $$f_n=1-1/2+1/3-1/4+\dots+\frac{(-1)^{n-1}}{n}$$ is a Cauchy sequence. 
My attempt:
Consider $$|f_{2m}-f_m| = \left| \frac{(-1)^m}{m+1}+\frac{(-1)^{m+1}}{m+2}+\dots+\frac{(-1)^{2m-1}}{2m} \right| \le \frac{1}{m+1}+\frac{1}{m+2}+\ldots+\frac{1}{2m}$$
But this is going in the direction of proving that the given sequence is not a Cauchy sequence.  Where am I doing it wrong?
 A: $$m\geqslant n\implies\left|f_m-f_n\right|\leqslant\frac1{n+1}$$
A: What's going wrong is that you are using the triangle equality too much.  This is an alternating series which does not converge absolutely so if you lose the signs of course it will not converge.  What I suggest trying is to group the terms in the partial sums "by two" so that all the terms are positive (or, all the terms are negative) and then the series will look more like the series for summing $\frac{1}{n^2}$.  Of course then you have to show this series converges but this is done in many textbooks, e.g. Rudin.
Of course you could apply the alternating series test to see that it converges and is thus Cauchy but I assume you want something more explicit.
A: $|f_{m+p}-f_m| = \left| \frac{(-1)^m}{m+1}+\frac{(-1)^{m+1}}{m+2}+\dots+\frac{(-1)^{m+p-1}}{m+p} \right| = \left| \frac{1}{m+1}+\frac{(-1)^{1}}{m+2}+\dots+\frac{(-1)^{p-1}}{m+p} \right| $ 
If $p$ is odd, then $|f_{m+p}-f_m| =\left| \frac{1}{m+1}+\frac{(-1)^{1}}{m+2}+\dots+\frac{(-1)^{p-1}}{m+p} \right|=\left| \bigg(\frac{1}{m+1}-\frac{1}{m+2}\bigg)+\bigg(\frac{1}{m+3}-\frac{1}{m+4}\bigg)+\dots+\bigg(\frac{1}{m+p-2} -\frac{1}{m+p-1}\bigg)+\frac{1}{m+p}\right|=\bigg(\frac{1}{m+1}-\frac{1}{m+2}\bigg)+\bigg(\frac{1}{m+3}-\frac{1}{m+4}\bigg)+\dots+\bigg(\frac{1}{m+p-2} -\frac{1}{m+p-1}\bigg)+\frac{1}{m+p}=\frac{1}{m+1}-\bigg(\frac{1}{m+2}-\frac{1}{m+3}\bigg)-\dots-\bigg(\frac{1}{m+p-1} -\frac{1}{m+p}\bigg)< \frac{1}{m+1} $
(because all the terms except the first are all negative)
If $p$ is even, then$|f_{m+p}-f_m| =\left| \frac{1}{m+1}+\frac{(-1)^{1}}{m+2}+\dots+\frac{(-1)^{p-1}}{m+p} \right|=\left| \bigg(\frac{1}{m+1}-\frac{1}{m+2}\bigg)+\bigg(\frac{1}{m+3}-\frac{1}{m+4}\bigg)+\dots+\bigg(\frac{1}{m+p-1} -\frac{1}{m+p}\bigg)\right|=\bigg(\frac{1}{m+1}-\frac{1}{m+2}\bigg)+\bigg(\frac{1}{m+3}-\frac{1}{m+4}\bigg)+\dots+\bigg(\frac{1}{m+p-1} -\frac{1}{m+p}\bigg)=\frac{1}{m+1}-\bigg(\frac{1}{m+2}-\frac{1}{m+3}\bigg)-\dots-\bigg(\frac{1}{m+p-2} -\frac{1}{m+p-1}\bigg)-\frac{1}{m+p}< \frac{1}{m+1} $
So for any $m, p>0, |f_{m+p}-f_m|\leq \frac{1}{m+1},$ this implies that $f_m$ is Cauchy sequence.
A: One answer was not written here down. It is written in a way called "Mathematics made difficult".
Consider the infinite series $\sum \frac{(-1)^{n-1}}{n}$. This is a convergent series as it is an alternating series and its $n$-th term $\frac{1}{n}$ converges to $0$ as $n \rightarrow \infty$.
$f_n = n$-th partial sum $= 1 - \frac{1}{2} + \frac{1}{3} + \dots + \frac{(-1)^{(n-1)}}{n}$. The sequence $\{f_n\}$ is a convergent sequence of real numbers. Hence it is Cauchy sequence.
