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

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

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sum from n=1 to infinity of 1/(n(n-1))

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

1

Solution steps

n = 2 \sum \infty \frac{1}{n ( n - 1 )}

Apply Telescoping Series Test:

1

= 1

n = 1 \sum \infty \frac{1}{5 ^{n} - 1}

n = 0 \sum \infty (- \frac{4}{5})^{n}

n = 2 \sum \infty (- 0.35)^{n}

n = 0 \sum \infty n^{\frac{1}{n}}

n = 1 \sum \infty (\frac{3 π}{4})^{n}

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