What is the sum from n=5 to infinity of 1/(n^2-n) ?

The sum from n=5 to infinity of 1/(n^2-n) is 1/4

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sum from n=5 to infinity of 1/(n^2-n)

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

\frac{1}{4}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

n = 1 \sum \infty \frac{2}{7 ^{n + 1}}

n = 1 \sum \infty \frac{1}{8 n + 1}

n = 1 \sum \infty 4 ln (\frac{n}{n + 1})

n = 0 \sum \infty (\frac{1}{12})^{n}

n = 4 \sum \infty ln (\frac{n}{2})

What is the sum from n=5 to infinity of 1/(n^2-n) ?
The sum from n=5 to infinity of 1/(n^2-n) is 1/4