1. ## Determinate the sums

Hi all i need to determinate the following sums:
$\sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}$
and
$\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$

How do i approach this? Thanx in advance

2. Originally Posted by Zaph
Hi all i need to determinate the following sums:
$\sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}$
and
$\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$

How do i approach this? Thanx in advance
Start by substituting the partial fraction decomposition of $\frac{1}{n(n+1)}$.

3. Originally Posted by mr fantastic
Start by substituting the partial fraction decomposition of $\frac{1}{n(n+1)}$.
Well that gives me $\frac{1}{n} \frac{1}{(n+1)}$

And then what?

4. Originally Posted by Zaph
Well that gives me $\frac{1}{n} \frac{1}{(n+1)}$

And then what?
Do you understand what is meant by partial fraction decomposition? $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. I suggest you now write out the first few terms using this and see what happens. It also might help to review your classnotes or textbook. Telescoping series would be a good place to start.

5. Originally Posted by mr fantastic
Do you understand what is meant by partial fraction decomposition? $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. I suggest you now write out the first few terms using this and see what happens. It also might help to review your classnotes or textbook. Telescoping series would be a good place to start.
Okay..
I think i found a pattern in the first few terms, as i get

$1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...$ so that means $1 - \frac{1}{2} + \frac{1}{2}..$ cancels eachother out meaning the sum is 1.

Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?

6. Originally Posted by Zaph
Okay..
I think i found a pattern in the first few terms, as i get

$1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...$ so that means $1 - \frac{1}{2} + \frac{1}{2}..$ cancels eachother out meaning the sum is 1.

Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
EDIT: The alternating series is wrong, gonna calc some more..

7. Originally Posted by Zaph
Okay..
I think i found a pattern in the first few terms, as i get

$1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...$ so that means $1 - \frac{1}{2} + \frac{1}{2}..$ cancels eachother out meaning the sum is 1.

Then for the alternating series, it gets timed by -1, and by the same idea it sums to -1?
Not all terms are multiplied by -1, only where n is odd... You will have an alternating series that is NOT telescopic.

Have a look at the series

$\sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = -\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) - \left(\frac{1}{3} - \frac{1}{4}\right)$
$+ \left(\frac{1}{4} - \frac{1}{5}\right) - \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) - \dots + \dots$

$= -1 + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots$

$= -1 + \frac{2}{1} - \frac{2}{1} + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots$

$= 1 - 2\sum_{n = 1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\right]$

This infinite sum is the alternating harmonic series, which is known to converge to $\ln{2}$.

So we finally have

$\sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = 1 - 2\ln{2}$.