Sum of Sequence of Products of Consecutive Reciprocals/Corollary

Corollary to Sum of Sequence of Products of Consecutive Reciprocals

 * $\displaystyle \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = 1$

Proof
From Sum of Sequence of Products of Consecutive Reciprocals:
 * $\displaystyle \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1} = 1 - \frac 1 {n + 1}$

We have that:
 * $\dfrac 1 {n + 1} < \dfrac 1 n$

and that $\sequence {\dfrac 1 n}$ is a basic null sequence.

Thus by the Squeeze Theorem:
 * $\displaystyle \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = 1$