What is the sum from n=0 to infinity of 1/(n^2-2n) ?

The sum from n=0 to infinity of 1/(n^2-2n) is converges

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sum from n=0 to infinity of 1/(n^2-2n)

n = 0 \sum \infty \frac{1}{n ^{2} - 2 n}

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

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

n = 1 \sum \infty e^{- \frac{n}{6}}

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

n = 4 \sum \infty \frac{2}{5 n ^{3}}

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

What is the sum from n=0 to infinity of 1/(n^2-2n) ?
The sum from n=0 to infinity of 1/(n^2-2n) is converges