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

The sum from n=1 to infinity of arccos(1/n)-arccos(1/(n+1)) is -pi/2

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

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

- \frac{π}{2}

Solution steps

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

Expand telescoping series:

- \frac{π}{2}

= - \frac{π}{2}

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

n = 1 \sum \infty 4 (\frac{3}{4})^{n - 1}

n = 1 \sum \infty \frac{2}{5 n - 1}

n = 1 \sum \infty \frac{8}{n ( n + 2 )}

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

What is the sum from n=1 to infinity of arccos(1/n)-arccos(1/(n+1)) ?
The sum from n=1 to infinity of arccos(1/n)-arccos(1/(n+1)) is -pi/2