What is the sum from k=1 to infinity of 4/(k(k+2)) ?

The sum from k=1 to infinity of 4/(k(k+2)) is 3

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sum from k=1 to infinity of 4/(k(k+2))

k = 1 \sum \infty \frac{4}{k ( k + 2 )}

3

Solution steps

k = 1 \sum \infty \frac{4}{k ( k + 2 )}

Apply the constant multiplication rule:

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

= 4 \cdot k = 1 \sum \infty \frac{1}{k ( k + 2 )}

Take the partial fraction of

\frac{1}{k ( k + 2 )} : \frac{1}{2 k} - \frac{1}{2 ( k + 2 )}

= 4 \cdot k = 1 \sum \infty \frac{1}{2 k} - \frac{1}{2 ( k + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

4 \cdot \frac{3}{4} : 3

= 3

n = 1 \sum \infty 7 (0.5)^{n - 1}

n = 2 \sum \infty \frac{4}{5 ^{n}}

n = 0 \sum \infty \frac{5}{1 + 9 n ^{2}}

n = 0 \sum \infty ln (4 n + 1)

n = 1 \sum \infty n \cdot e^{- n}

What is the sum from k=1 to infinity of 4/(k(k+2)) ?
The sum from k=1 to infinity of 4/(k(k+2)) is 3