What is the sum from n=1 to infinity of 6/(n^2+2n) ?

The sum from n=1 to infinity of 6/(n^2+2n) is 9/2

Popular >

sum from n=1 to infinity of 6/(n^2+2n)

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

\frac{9}{2}

Solution steps

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

Apply the constant multiplication rule:

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

= 6 \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 )}

= 6 \cdot n = 1 \sum \infty \frac{1}{2 n} - \frac{1}{2 ( n + 2 )}

Expand telescoping series:

\frac{3}{4}

Simplify

6 \cdot \frac{3}{4} : \frac{9}{2}

= \frac{9}{2}

\int_{1}^{10} \frac{21}{x ^{3}} d x

x \to - 1 lim (\frac{4}{( x + 1 ) ^{2}})

\frac{d}{d x} (sin^{2} (\frac{x ^{2} + 1}{x}))

x \to 4 - lim (\frac{1}{4 - x})

\frac{11}{( 2 x ^{3} - 5 ) ^{4}}

What is the sum from n=1 to infinity of 6/(n^2+2n) ?
The sum from n=1 to infinity of 6/(n^2+2n) is 9/2