What is the sum from n=1 to infinity of 1/(n*(n+1)) ?

The sum from n=1 to infinity of 1/(n*(n+1)) is 1

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sum from n=1 to infinity of 1/(n*(n+1))

n = 1 \sum \infty \frac{1}{n \cdot ( n + 1 )}

1

Solution steps

n = 1 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

1

= 1

\int \frac{1 + x}{- x x} d x

\int_{0}^{1} - (x^{2} - x) d x

x \to 3 lim (- 7 - x)

\int x e^{2 x} d x

y^{^{''''}} - 4 y^{^{''}} = 0

What is the sum from n=1 to infinity of 1/(n*(n+1)) ?
The sum from n=1 to infinity of 1/(n*(n+1)) is 1