What is the sum from n=3 to infinity of 6/(n(n-2)) ?

The sum from n=3 to infinity of 6/(n(n-2)) is 9/2

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sum from n=3 to infinity of 6/(n(n-2))

n = 3 \sum \infty \frac{6}{n ( n - 2 )}

\frac{9}{2}

Solution steps

n = 3 \sum \infty \frac{6}{n ( n - 2 )}

Apply the constant multiplication rule:

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

= 6 \cdot n = 3 \sum \infty \frac{1}{n ( n - 2 )}

Take the partial fraction of

\frac{1}{n ( n - 2 )} : - \frac{1}{2 n} + \frac{1}{2 ( n - 2 )}

= 6 \cdot n = 3 \sum \infty - \frac{1}{2 n} + \frac{1}{2 ( n - 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

\frac{x ^{2} - 7 x + 3}{x ^{2} + 7}

y^{^{''}} + 4 y^{^{'}} + 5 y = 15 x + 5 e^{- x}

\frac{d}{d x} (e^{(x + y)^{2}})

x \to 4 + lim (- 3)

f (x) = \frac{2 x}{x ^{2} + 1}, 1, at x = 1

What is the sum from n=3 to infinity of 6/(n(n-2)) ?
The sum from n=3 to infinity of 6/(n(n-2)) is 9/2