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

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

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

n = 1 \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 = 1 \sum \infty tan^{n} (\frac{π}{6})

n = 1 \sum \infty 9 n^{2} e^{- n^{3}}

n = 0 \sum \infty (1 + \frac{1}{2 n})^{n}

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

n = \frac{π}{2} \sum \infty sin (n)

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