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 ( n + 1 ) )}

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

n = 0 \sum \infty 1 + (- 1)^{n}

n = 0 \sum \infty ln (5 - n)

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

n = 1 \sum \infty \frac{2}{4 ^{n}}

n = 0 \sum \infty \frac{6 ^{n}}{n !}

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