By expressing the

*r*th term of the series $\displaystyle \sum^n_{r=1} \frac{(-1)^{r+1}(2r+1)}{r(r+1)}$ in partial fractions, or otherwise, find the sum of the series.

If this sum is denoted by $\displaystyle S_n$ prove that for all odd values of n, $\displaystyle S_nS_{n+1}=1$

I found the partial fractions to be:

$\displaystyle \sum^n_{n=1}(\frac{-1}{r}+\frac{1}{r+1})$

and the sum of the series:

$\displaystyle \frac{1}{n+1}-1$

but the sum is given as $\displaystyle 1+\frac{(-1)^{n+1}}{n+1}$

Thanks!