What is the sum from n=2 to infinity of 9/(n(n+2)) ?

The sum from n=2 to infinity of 9/(n(n+2)) is converges

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sum from n=2 to infinity of 9/(n(n+2))

n = 2 \sum \infty \frac{9}{n ( n + 2 )}

converges

Solution steps

n = 2 \sum \infty \frac{9}{n ( n + 2 )}

Apply the constant multiplication rule:

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

= 9 \cdot n = 2 \sum \infty \frac{1}{n ( n + 2 )}

Apply Series Comparison Test:

converges

= 9 converges

= converges

\frac{d}{d x} (\frac{x}{1 + x ^{2} + x ^{4}})

\frac{d}{d x} (- 7 x^{3} - \frac{2}{x ^{3}} + 7 x)

\int \frac{1}{25 x ^{2} - 81} d x

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sin^{3} (cos (8 x))

What is the sum from n=2 to infinity of 9/(n(n+2)) ?
The sum from n=2 to infinity of 9/(n(n+2)) is converges