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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

y^{^{'}} - 5 y = 2 e^{5 x}

x^{3} - 4 x

\frac{d}{d x} (7 - x)

\int \frac{x ^{2}}{4 x ^{3} + 5} d x

f (x) = 5 x^{3} ln (x)

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