Application of Cauchy's convergence criterion Use Cauchy's convergence criterion to prove convergence of $x_n$
$$x_n=1-\frac{1}{2}+\frac{1}{3}-\cdots+(-1)^{n+1}\frac{1}{n}$$
as far as I am concerned,supposing that $m>n$, it's apparently warranted that $|x_m-x_n|=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}$
but, furthermore, it was showed on the textbook that $\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k}<\dfrac{1}{n+1}$
I wonder how can we approach the last step

My post will going to aim and clarify that inequality in your textbook by using some elementary approach. First of all, notice that each term in your series that is set on an odd position is positive. Let's fix n as an even number and fix an even number of terms of your series and then we get: 
$$S=\sum_{k=n+1}^m (-1)^{k+1}\dfrac{1}{k} = \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3}-\frac{1}{n+4}+\frac{1}{n+5}-\frac{1}{n+6}+\cdots $$
$$= \frac{1}{(n+1)(n+2)}+\frac{1}{(n+3)(n+4)}+\frac{1}{(n+5)(n+6)}+ ...$$
Here we are going to use an inequality with a sum that telescopes (on the right side):
$$S< \frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+3)(n+4)}+ ...=$$
$$\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+2}-\frac{1}{n+3}+\frac{1}{n+3}-\frac{1}{n+4}+...<\frac{1}{n+1}.$$
Notice that if you have an odd number of terms, then when you write down the
inequality and
take out from both sides the term $\frac1{n+1}$ then on the left side you get an even number of terms and you can group them such that every pair is a negative sum. Moreover, if your sum starts with a negative number, then the sum is obviously negative because the terms of the sum decrease. 
Therefore, the inequality holds.
The proof is complete.  
