# 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

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 6.3$ - 1992: Larry C. Andrews:
*Special Functions of Mathematics for Engineers*... (previous) ... (next): $\S 1.2$: Infinite Series of Constants: Example $1$