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

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

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

k = 3 \sum \infty \frac{1}{k ^{2} - k}

\frac{1}{2}

Solution steps

k = 3 \sum \infty \frac{1}{k ^{2} - k}

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

n = 1 \sum \infty n \cdot (\frac{1}{3})^{n}

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

n = 1 \sum \infty \frac{n ^{2}}{9 ^{n}}

k = 2 \sum \infty \frac{k ^{e}}{k ^{π}}

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

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