What is the sum from n=1 to infinity of 2/(n*(n+1)) ?

The sum from n=1 to infinity of 2/(n*(n+1)) is 2

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sum from n=1 to infinity of 2/(n*(n+1))

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

2

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

1

= 2 \cdot 1

Simplify

= 2

\int w^{2} d w

g (x) = (4 x^{3} + 6 x^{2})^{5}

\frac{d}{d θ} (2 sin^{2} (θ))

\frac{d}{d x} ((a x - b) (c x^{2}))

x \to - \infty lim (\frac{( x ^{- 1} + x ^{- 4} )}{x ^{- 2} - x ^{- 3}})

What is the sum from n=1 to infinity of 2/(n*(n+1)) ?
The sum from n=1 to infinity of 2/(n*(n+1)) is 2