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

The sum from n=4 to infinity of 1/(n^2-n) is 1/3

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

n = 4 \sum \infty \frac{1}{n ^{2} - n}

\frac{1}{3}

Solution steps

n = 4 \sum \infty \frac{1}{n ^{2} - n}

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

n = 0 \sum \infty (- 9)^{n}

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

n = 0 \sum \infty 8 e^{- n}

n = 2 \sum \infty (- 0.1)^{n}

n = 0 \sum \infty (\frac{4}{3})^{- n}

What is the sum from n=4 to infinity of 1/(n^2-n) ?
The sum from n=4 to infinity of 1/(n^2-n) is 1/3