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

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

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

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

- \frac{1}{2}

Decimal

- 0.5

Solution steps

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

Expand

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

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

Apply the constant multiplication rule:

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

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

Apply Telescoping Series Test:

\frac{1}{2}

= - \frac{1}{2}

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