Sum of series using method of differences

Question

How do I show that $$\sum_{k=1}^n\left(\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+2}\right) =\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}.$$ using the method of differences?

Answer 1

I guess that there is a typo in the formula. It should be $$\sum_{k=1}^n\left(\frac{1}{k}-\frac{2}{k+1}+\frac{1}{k+2}\right) =\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}.$$

I do not know what you mean by "method of differences". Anyway you can show the identity in the following way.

Let $H_n=\sum_{k=1}^n\frac{1}{k}$. Then $$\sum_{k=1}^n\left(\frac{1}{k}-\frac{2}{k+1}+\frac{1}{k+2}\right)= \sum_{k=1}^n\frac{1}{k}-2\sum_{k=1}^n\frac{1}{k+1}+\sum_{k=1}^n\frac{1}{k+2}\\ =\sum_{k=1}^n\frac{1}{k}-2\sum_{k=2}^{n+1}\frac{1}{k}+ \sum_{k=3}^{n+2}\frac{1}{k}\\ =H_n-2\left(H_n-1+\frac{1}{n+1}\right)+\left(H_n-1-\frac{1}{2}+\frac{1}{n+1}+\frac{1}{n+2}\right)\\ =\frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2}.$$