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}

\frac{d}{d x} (\frac{4 x}{( x ^{2} + 1 )})

\int ln (17 t) d t

\frac{d}{d x} (\frac{7}{5} sin^{2} (2 x))

\int sec^{2} (2 x - 1) d x

y (x - y - 2) d x + x (y - x + 4) d y = 0

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