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

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

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

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

\frac{1}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

1

= \frac{1}{2} \cdot 1

Simplify

= \frac{1}{2}

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y = x^{2}, y = 6 - x

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f (x) = \frac{1}{x}, at x = a

\frac{d}{d p} (\frac{p - 1}{p ^{2} + 4})

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