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

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