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

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

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

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

\frac{3}{2}

Solution steps

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

Apply the constant multiplication rule:

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

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

Take the partial fraction of

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

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

Expand telescoping series:

\frac{3}{4}

Simplify

2 \cdot \frac{3}{4} : \frac{3}{2}

= \frac{3}{2}

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

n = 1 \sum \infty \frac{n + 6}{n ^{2}}

n = 1 \sum \infty \frac{2 n}{n ^{2} + 4}

k = 2 \sum \infty \frac{k}{ln ( k )}

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

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