What is the sum from n=2 to infinity of 2/(n^2-n) ?

The sum from n=2 to infinity of 2/(n^2-n) is 2

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sum from n=2 to infinity of 2/(n^2-n)

n = 2 \sum \infty \frac{2}{n ^{2} - n}

2

Solution steps

n = 2 \sum \infty \frac{2}{n ^{2} - n}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 2 \sum \infty \frac{1}{n ^{2} - n}

Apply Telescoping Series Test:

1

= 2 \cdot 1

Simplify

= 2

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What is the sum from n=2 to infinity of 2/(n^2-n) ?
The sum from n=2 to infinity of 2/(n^2-n) is 2