Originally Posted by

**aukie** Hello Tonio

In working out why the absolute value can be ommitted and what you meant by trivial. I seen an argument that doesn't require induction. If you're interested: Assuming m = n + k,

$\displaystyle \frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k}$

Group the terms as follows (for even k)

$\displaystyle \frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - (\frac{1}{n+(k-1)} - \frac{1}{n+k})$

(and for odd k)

$\displaystyle \frac{1}{n+1} - (\frac{1}{n+2} - \frac{1}{n+3}) - \text{...} - ( \frac{1}{n+k})$

Then we have a series of positive terms subtracted from $\displaystyle \frac{1}{n+1}$ so

$\displaystyle \frac{1}{n+1} - \frac{1}{n+2} + \text{...} \pm \frac{1}{n+k} < \frac{1}{n+1} < \frac{1}{n}$

Thanks for your input