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

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

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

n = 1 \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}

\frac{\partial}{\partial x} (r \frac{g ( r , x )}{2 h})

\int \frac{1}{4 x + x ^{2}} d x

y^{2} d t + t^{2} d y = 0

\frac{d}{d θ} (4 cos (3 θ))

\int \frac{1}{1 ^{2} + u ^{2}} d u

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