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 1/4

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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}{4}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

\int_{2}^{4} (x^{4} - 1) d x

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

x \to 0 lim ((1 + 4 x)^{\frac{6}{x}})

\frac{\partial}{\partial y} (3 x^{2} y + 5 x y^{3})

\frac{( 3 - x )}{4 x}

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 1/4