Sum of Sequence of Products of Consecutive Reciprocals/Corollary

Corollary to Sum of Sequence of Products of Consecutive Reciprocals

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

Proof

From Sum of Sequence of Products of Consecutive Reciprocals:

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

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:

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

$\blacksquare$

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 (2nd ed.) ... (previous) ... (next): $\S 1.2$: Infinite Series of Constants: Example $1$