What is the sum from n=5 to infinity of 1/(n(n+1)) ?

The sum from n=5 to infinity of 1/(n(n+1)) is 1/5

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sum from n=5 to infinity of 1/(n(n+1))

n = 5 \sum \infty \frac{1}{n ( n + 1 )}

\frac{1}{5}

Solution steps

n = 5 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{5}

= \frac{1}{5}

\frac{d y}{d x} = sin (5 x)

\int 5 x - 1 d x

\frac{d P}{d t} + 2 tP = P + 8 t - 4

\int - \frac{3}{x ^{2}} d x

\frac{\partial}{\partial y} (ln (x^{10} y))

What is the sum from n=5 to infinity of 1/(n(n+1)) ?
The sum from n=5 to infinity of 1/(n(n+1)) is 1/5