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

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

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

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

- 1

Solution steps

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

Apply Telescoping Series Test:

- 1

= - 1

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

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

n = 2 \sum \infty (\frac{1}{3})^{2 n}

n = 0 \sum \infty \frac{1}{n ^{e}}

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

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