What is the sum from n=7 to infinity of 1/(n(n+1)) ?

The sum from n=7 to infinity of 1/(n(n+1)) is 1/7

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sum from n=7 to infinity of 1/(n(n+1))

n = 7 \sum \infty \frac{1}{n ( n + 1 )}

\frac{1}{7}

Solution steps

n = 7 \sum \infty \frac{1}{n ( n + 1 )}

Apply Telescoping Series Test:

\frac{1}{7}

= \frac{1}{7}

n = 1 \sum \infty \frac{7}{n ^{2} + 16}

n = 0 \sum \infty \frac{n}{7 n ^{3} + 1}

n = 0 \sum \infty ln (2 - n)

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

n = 4 \sum \infty \frac{1}{n ^{2} - 4 n + 3}

What is the sum from n=7 to infinity of 1/(n(n+1)) ?
The sum from n=7 to infinity of 1/(n(n+1)) is 1/7