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

The sum from k=1 to infinity of 1/k-1/(k+2) is 3/2

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

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

\frac{3}{2}

Solution steps

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

Expand telescoping series:

\frac{3}{2}

= \frac{3}{2}

n = 1 \sum \infty 6 ln (\frac{n}{n + 1})

k = 0 \sum \infty \frac{k}{2 k + 1}

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

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

n = 1 \sum \infty e^{- \frac{n}{6}}

What is the sum from k=1 to infinity of 1/k-1/(k+2) ?
The sum from k=1 to infinity of 1/k-1/(k+2) is 3/2