What is the sum from n=5 to infinity of 1/n-1/(n+1) ?

The sum from n=5 to infinity of 1/n-1/(n+1) is 1/5

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sum from n=5 to infinity of 1/n-1/(n+1)

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

\frac{1}{5}

Solution steps

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

Apply Telescoping Series Test:

\frac{1}{5}

= \frac{1}{5}

n = 0 \sum \infty (\frac{1}{2 - i})^{n}

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

n = 1 \sum \infty tan^{n} (\frac{π}{6})

n = 1 \sum \infty 9 n^{2} e^{- n^{3}}

n = 0 \sum \infty (1 + \frac{1}{2 n})^{n}

What is the sum from n=5 to infinity of 1/n-1/(n+1) ?
The sum from n=5 to infinity of 1/n-1/(n+1) is 1/5