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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Integral Test:

converges

= 3 converges

= converges

(5 x + 7)^{2}

\int sin (4 t) cos (4 t) d t

\int \frac{1}{x ( x + 4 )} d x

\frac{d ^{2}}{d x ^{2}} (e^{\frac{1}{x}})

\frac{d y}{d x} = \frac{y}{x ^{2}}, y (1) = 2

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