- Changing the indices in the infinite sum, and using the definition of the falling factorial powers yields $$\sum_{x=1}^\infty\frac{1}{x(x+1)}=\sum_{x=0}^\infty\frac{1}{(x+1)(x+2)}=\sum_{x=0}^\infty x^{\underline{-2}}.$$
- We have to evaluate the limit of the partial sums $$S_n:=\sum_{x=0}^n x^{\underline{-2}}$$ for $n\to\infty.$
- Applying the fundamental theorem of the difference calculus we get $$S_n=\frac{x^{\underline {-1}}}{-1}\;\Rule{1px}{4ex}{2ex}^{n+1}_{0}=-\frac{1}{x+1}\;\Rule{1px}{4ex}{2ex}^{n+1}_{0}=1-\frac{1}{n+2}.$$
- And $$\lim_{n\to\infty} S_n=1.$$

| | | | created: 2020-05-17 11:02:02 | modified: 2020-05-17 11:09:59 | by: *bookofproofs* | references: [8404]

[8404] **Miller, Kenneth S.**: “An Introduction to the Calculus of Finite Differences And Difference Equations”, Dover Publications, Inc, 1960