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

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

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

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

\frac{1}{4}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{4}

= \frac{1}{4}

n = 0 \sum \infty (2)^{n}

n = 1 \sum \infty \frac{3 ^{n}}{n + 1}

k = 1 \sum \infty ln (\frac{k}{3 k + 1})

n = 0 \sum \infty \frac{n}{8 n + 1}

n = 0 \sum \infty sin (\frac{n}{2})

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