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

The sum from n=2 to infinity of 3/(n(n+2)) is converges

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 3 converges

= converges

\frac{d y}{d x}, y = 2 x^{2} + 5 y x

\int x^{2} \cdot arctan (x) d x

\int_{1}^{x} t^{2} d t

x y - 8 y^{2} = 4, (18, 2)

5 + 2 t cos (2 t)

What is the sum from n=2 to infinity of 3/(n(n+2)) ?
The sum from n=2 to infinity of 3/(n(n+2)) is converges