Good evening all,
I am faced with this dilemma, and I am hoping someone can help me out.
The series is,
$$\sum_{n=9}^{\infty}\frac{1}{n(n-1)}
$$
I have figured out the sum to be $\frac{1}{8}$ but I cannot seem to get the expansion as a telescoping series correct
I have tried $\frac{1}{N}$ – $\frac{1}{N-1}$ and $\frac{1}{N+1}$ – $\frac{1}{N}$ but neither work..
If anyone has any idea on how to write the series as a telescoping series that would be greatly appreciated. Any hints or tips are welcome!
Best Answer
$$\sum_{n=9}^{\infty}\frac{1}{n(n-1)}=\lim_{N\to\infty}\sum_{n=9}^{N}\left(\frac{1}{n-1}-\frac1n\right)=\lim_{N\to\infty}\frac18-\frac1N=\frac18 $$