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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= 2 converges

= converges

\int - 5 csc^{2} (x) d x

x \to - 3 + lim (\frac{x + 4}{x + 3})

x \to + (- 4) lim (∣ x + 4 ∣)

y = 3 x - 4, (2, 5)

3 x \frac{d y}{d x} + y = x

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