What is the sum from n=0 to infinity of 1/(n*(n+1)) ?

The sum from n=0 to infinity of 1/(n*(n+1)) is 1

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sum from n=0 to infinity of 1/(n*(n+1))

n = 0 \sum \infty \frac{1}{n \cdot ( n + 1 )}

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

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

n = 1 \sum \infty \frac{1}{3 ^{n} n}

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

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

n = 1 \sum \infty \frac{2}{3 n + 8}

What is the sum from n=0 to infinity of 1/(n*(n+1)) ?
The sum from n=0 to infinity of 1/(n*(n+1)) is 1