What is the sum from n=5 to infinity of 3/(n^2-n) ?

The sum from n=5 to infinity of 3/(n^2-n) is 3/4

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sum from n=5 to infinity of 3/(n^2-n)

n = 5 \sum \infty \frac{3}{n ^{2} - n}

\frac{3}{4}

Solution steps

n = 5 \sum \infty \frac{3}{n ^{2} - n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 5 \sum \infty \frac{1}{n ^{2} - n}

Apply Telescoping Series Test:

\frac{1}{4}

= 3 \cdot \frac{1}{4}

Simplify

3 \cdot \frac{1}{4} : \frac{3}{4}

= \frac{3}{4}

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What is the sum from n=5 to infinity of 3/(n^2-n) ?
The sum from n=5 to infinity of 3/(n^2-n) is 3/4