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

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

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

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

\frac{1}{7}

Solution steps

n = 7 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

Apply Telescoping Series Test:

\frac{1}{7}

= \frac{1}{7}

i = 1 \sum \infty (\frac{1}{2})^{i}

n = 0 \sum \infty \frac{1}{8 - n}

n = 1 \sum \infty 9 (0.1)^{n - 1}

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

n = 1 \sum \infty 3 (0.3)^{n - 1}

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