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

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

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

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

\frac{1}{3}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{3}

= \frac{1}{3}

n = 1 \sum \infty \frac{e ^{n}}{4 ^{n}}

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

x = 2 \sum \infty \frac{1}{x ln ( x )}

n = 1 \sum \infty cos (n \cdot π)

n = 5 \sum \infty \frac{1}{n}

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