Show that sequence is a Cauchy sequence Prove that given sequence $$\langle f_n\rangle =1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{(-1)^{n-1}}{n}$$ 
is a Cauchy sequence 
My attempt : 
$|f_{n}-f_{m}|=\Biggl|\dfrac{(-1)^{m}}{m+1}+\dfrac{(-1)^{m+1}}{m+2}\cdots\dots+\dfrac{(-1)^{n-1}}{n}\Biggr|$
using $ m+1>m \implies \dfrac{1}{m+1}<\dfrac{1}{m} $
$|f_{n}-f_{m}|\le \dfrac{1}{m}+\dfrac{1}{m}+\dfrac{1}{m}\cdots\cdots\dfrac{1}{m}$
$|f_{n}-f_{m}|\le\dfrac{n-m}{m}$
I don't know if I am proceeding correctly or if I am, how to proceed further, any hint would be really helpful .
 A: Hint :
$$\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{(2n-1)2n}\leq \frac{1}{(n-1)n}= \frac{1}{n-1}-\frac{1}{n} $$
A: If you ignore the signs of the terms,
the result diverges.
So you can't do that.
$f_n
=\sum_{k=1}^n \dfrac{(-1)^k}{k}
$
so,
if $n > m$,
$f_n-f_m
=\sum_{k=m+1}^n \dfrac{(-1)^k}{k}
=\sum_{k=1}^{n-m} \dfrac{(-1)^{k+m}}{k+m}
=(-1)^m\sum_{k=1}^{n-m} \dfrac{(-1)^{k}}{k+m}
$.
If
$n-m$ is even,
so $n-m = 2j$,
then
$\begin{array}\\
f_n-f_m
&=(-1)^m\sum_{k=1}^{2j} \dfrac{(-1)^{k}}{k+m}\\
&=(-1)^m\sum_{k=1}^{j} \left(\dfrac{(-1)^{2k-1}}{2k-1+m}+\dfrac{(-1)^{2k}}{2k+m}\right)\\
&=(-1)^m\sum_{k=1}^{j} (-1)^{2k-1}\left(\dfrac{-1}{2k-1+m}+\dfrac{1}{2k+m}\right)\\
&=(-1)^m\sum_{k=1}^{j} (-1)^{2k-1}\left(\dfrac{(2k-1+m)-(2k+m)}{(2k-1+m)(2k+m)}\right)\\
&=(-1)^{m+1}\sum_{k=1}^{j} \left(\dfrac{-1}{(2k-1+m)(2k+m)}\right)\\
&=(-1)^{m}\sum_{k=1}^{j} \left(\dfrac{1}{(2k-1+m)(2k+m)}\right)\\
\text{so}\\
|f_n-f_m|
&=\sum_{k=1}^{j} \left(\dfrac{1}{(2k-1+m)(2k+m)}\right)\\
&=\sum_{k=1}^{j}\dfrac14 \left(\dfrac{1}{(k-\frac12+\frac{m}{2})(k+\frac{m}{2})}\right)\\
&\lt \dfrac14\sum_{k=1}^{j} \left(\dfrac{1}{(k-1+\frac{m}{2})(k+\frac{m}{2})}\right)
\quad\text{this is the sneaky part}\\
&\lt \dfrac14\sum_{k=1}^{j} \left(\dfrac{1}{k-1+\frac{m}{2}}-\dfrac{1}{k+\frac{m}{2}}\right)\\
&= \dfrac14 \left(\dfrac{1}{\frac{m}{2}}-\dfrac{1}{j+\frac{m}{2}}\right)\\
&= \dfrac12 \left(\dfrac{1}{m}-\dfrac{1}{2j+m}\right)\\
&= \dfrac12 \left(\dfrac{1}{m}-\dfrac{1}{n}\right)\\
&< \dfrac{1}{2m}\\
&\to 0 \text{ as } m \to \infty\\
\end{array}
$
If $n-m$ is odd,
the sum changes
by at most $\frac1{n}$
so it still goes to zero.
