What is the sum from n=1 to infinity of 3/(n(n+1)) ?

The sum from n=1 to infinity of 3/(n(n+1)) is 3

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sum from n=1 to infinity of 3/(n(n+1))

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

3

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

1

= 3 \cdot 1

Simplify

= 3

f (x) = x, g (x) = x, x = 0, 2

x y^{^{'}} = 8 y - 9 x, y (1) = 0, y (1) = 0

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\frac{d}{d x} ((2 x^{5} - 5 x^{3} + 7 x)^{4})

What is the sum from n=1 to infinity of 3/(n(n+1)) ?
The sum from n=1 to infinity of 3/(n(n+1)) is 3