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

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

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

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

- \frac{1}{2}

Solution steps

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

Expand

\frac{1}{n + 1} - \frac{1}{n} : - \frac{1}{n ^{2} + n}

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Telescoping Series Test:

\frac{1}{2}

= - \frac{1}{2}

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

n = 1 \sum \infty \frac{5}{( ( - 4 ) ^{n} )}

k = 0 \sum \infty (\frac{1}{10})^{k}

k = 1 \sum \infty 3 (- \frac{1}{5})^{3 k}

n = 0 \sum \infty - \frac{3}{( n ^{2} - 4 )}

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