What is the sum from n=4 to infinity of 1/((n)(n+1)) ?

The sum from n=4 to infinity of 1/((n)(n+1)) is 1/4

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sum from n=4 to infinity of 1/((n)(n+1))

n = 4 \sum \infty \frac{1}{( n ) ( n + 1 )}

\frac{1}{4}

Solution steps

n = 4 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

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

x \to 0 lim (\frac{1}{1 - e ^{x}})

n = 1 \sum \infty (cos (32))^{n}

\frac{\partial}{\partial x} (x^{3} - 9 x y + 7 y^{3})

\frac{d}{d x} (750 + 12 x - 0.8 x^{7})

What is the sum from n=4 to infinity of 1/((n)(n+1)) ?
The sum from n=4 to infinity of 1/((n)(n+1)) is 1/4