What is the sum from n=3 to infinity of 1/(n(n+1)) ?

The sum from n=3 to infinity of 1/(n(n+1)) is 1/3

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sum from n=3 to infinity of 1/(n(n+1))

n = 3 \sum \infty \frac{1}{n ( n + 1 )}

\frac{1}{3}

Solution steps

n = 3 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

\int \frac{a e ^{θ} + b}{a e ^{θ} - b} d θ

y = \frac{1}{1 + e ^{- x}}

\frac{\partial}{\partial x} (e^{2 x} sin (2 y))

x \to 2 lim (\frac{x ^{2} - 4}{x + 2})

2 x^{^{''}} - 3 x^{^{'}} + x = 0

What is the sum from n=3 to infinity of 1/(n(n+1)) ?
The sum from n=3 to infinity of 1/(n(n+1)) is 1/3