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 .
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.