Sum of Sequence of Products of Consecutive Reciprocals/Corollary

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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:

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

$\blacksquare$


Sources