Hi all i need to determinate the following sums:
$\displaystyle \sum_{n=1}^{\infty} (-1)^n\frac{1}{n(n+1)}$
and
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$
How do i approach this? Thanx in advance
Do you understand what is meant by partial fraction decomposition? $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$. 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
$\displaystyle 1 - \frac{1}{2} + \frac{1}{2}-\frac{1}{3} +...$ so that means $\displaystyle 1 - \frac{1}{2} + \frac{1}{2}..$ 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
$\displaystyle \sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = -\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) - \left(\frac{1}{3} - \frac{1}{4}\right)$
$\displaystyle + \left(\frac{1}{4} - \frac{1}{5}\right) - \left(\frac{1}{5} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{7}\right) - \dots + \dots$
$\displaystyle = -1 + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots$
$\displaystyle = -1 + \frac{2}{1} - \frac{2}{1} + \frac{2}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \dots - \dots$
$\displaystyle = 1 - 2\sum_{n = 1}^{\infty}\left[\frac{(-1)^{n-1}}{n}\right]$
This infinite sum is the alternating harmonic series, which is known to converge to $\displaystyle \ln{2}$.
So we finally have
$\displaystyle \sum_{n = 1}^{\infty}\left[(-1)^n\left(\frac{1}{n} - \frac{1}{n - 1}\right)\right] = 1 - 2\ln{2}$.