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 1/2

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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}{2}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

\int (y^{2} + 1) d y

h \to 0 lim (\frac{( 8 ^{h} - 1 )}{h})

t^{2} - 8 t + 19

n = 0 \sum \infty \frac{x ^{n}}{5 ^{n}}

\frac{\partial}{\partial x} (\frac{ln ^{^{'}} ( x ^{2} + 5 )}{y ^{2}})

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 1/2