What is the sum from n=0 to infinity of 2/(n^2+2n) ?

The sum from n=0 to infinity of 2/(n^2+2n) is 3/2

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sum from n=0 to infinity of 2/(n^2+2n)

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

\frac{3}{2}

Solution steps

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

Take the partial fraction of

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

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

Expand telescoping series:

\frac{3}{4}

Simplify

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

= \frac{3}{2}

\int \frac{x ^{2} + 3}{x ^{3} + 2 x} d x

\frac{d}{d x} (ln (3 x^{2} - y^{3} - 3))

\frac{d}{d x} (3^{x^{2} c o t (2 x)})

\frac{d}{d x} (\frac{8}{3 x})

\frac{x + 3}{x - 3}

What is the sum from n=0 to infinity of 2/(n^2+2n) ?
The sum from n=0 to infinity of 2/(n^2+2n) is 3/2