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

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

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sum from n=2 to infinity of 1/n-1/(n+1)

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

\frac{1}{2}

Solution steps

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

Simplify

\frac{1}{n} - \frac{1}{n + 1} : \frac{1}{n ( n + 1 )}

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

Apply Telescoping Series Test:

\frac{1}{2}

= \frac{1}{2}

n = 1 \sum \infty \frac{1 6 ^{n}}{n !}

k = 1 \sum \infty k e^{- 4 k^{2}}

n = 2 \sum \infty \frac{7}{n ln ( n )}

n = 0 \sum \infty \frac{n !}{40 0 ^{n}}

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

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