1. ## Sum of series

By expressing the rth 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!

2. Originally Posted by arze
By expressing the rth 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!

3. Ok this is the way I worked it out:
For the part $\displaystyle \frac{}{r}...\frac{}{r+1}$
When r=0, the other part of becomes $\displaystyle \frac{(-1)^{1}(1)}{1}$
hence $\displaystyle \frac{-1}{r}$
then the next part, r=-1, and the part becomes $\displaystyle \frac{(-1)^{0}(-1)}{-1}$
hence$\displaystyle \frac{1}{r+1}$
hence we get the partial fraction $\displaystyle -\frac{1}{r}+\frac{1}{r+1}$
Is this correct?
Thanks!

4. No, the partial fraction decomposition is

$\displaystyle \displaystyle \frac{(-1)^{r+1}(2r+1)}{r(r+1)} = (-1)^{r+1}(\frac{1}{r}+\frac{1}{r+1})$

Also here's how to get it

$\displaystyle \displaystyle \frac{(2r+1)}{r(r+1)} = \frac{A}{r}+\frac{B}{r+1} = \frac{(A+B)r+A}{r(r+1)}$

So you have $\displaystyle 2r+1 = (A+B)r+A$ is valid for all r, so let

r = 0, you get A = 1
r = -1/2, you get A/2-B/2 = 0 which implies A = B

Then just multiply the result by (-1)^{r+1} and then just do your sum.

5. Even more simple...

$\displaystyle \displaystyle \frac{2r+1}{r\ (r+1)} = \frac{2}{r} - \frac{1}{r\ (r+1)}= \frac{1}{r} + \frac{1}{r+1}$ (1)

... and now...

$\displaystyle \displaystyle \sum_{r=1}^{n} \frac{(-1)^{r+1}}{r} = 1 - \frac{1}{2} + \frac{1}{3} -...+ \frac{(-1)^{n+1}}{n}$ (2)

$\displaystyle \displaystyle \sum_{r=1}^{n} \frac{(-1)^{r+1}}{r+1} = \frac{1}{2} - \frac{1}{3} -...+ \frac{(-1)^{n+1}}{n+1}$ (3)

At this point You have to sum (2) and (3) toghether...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

6. $\displaystyle \dfrac{2r+1}{r(r+1)}=\dfrac{r+r+1}{r(r+1)}=\dfrac{ 1}{r+1}+\dfrac{1}{r}.$

that's actually even easier.