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

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

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

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

1

Solution steps

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

Apply Telescoping Series Test:

1

= 1

n = 0 \sum \infty \frac{6}{3 ^{n}}

n = 1 \sum \infty \frac{5}{3 n ^{3} + 1}

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

n = 1 \sum \infty 5 e^{- 0.5 n}

n = 0 \sum \infty \frac{7}{n + 3}

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