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

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

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

k = 3 \sum \infty \frac{2}{k ( k + 1 )}

\frac{2}{3}

Solution steps

k = 3 \sum \infty \frac{2}{k ( k + 1 )}

Apply the constant multiplication rule:

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

= 2 \cdot k = 3 \sum \infty \frac{1}{k ( k + 1 )}

Apply Telescoping Series Test:

\frac{1}{3}

= 2 \cdot \frac{1}{3}

Simplify

2 \cdot \frac{1}{3} : \frac{2}{3}

= \frac{2}{3}

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

k = 2 \sum \infty (\frac{3}{8})^{2 k}

n = 2 \sum \infty arctan (n)

n = 0 \sum \infty ln (1 + 5 n)

n = 0 \sum \infty 3 (\frac{1}{11})^{n}

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