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

The sum from n=1 to infinity of 3/(n^2+3n) is 11/6

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

n = 1 \sum \infty \frac{3}{n ^{2} + 3 n}

\frac{11}{6}

Solution steps

n = 1 \sum \infty \frac{3}{n ^{2} + 3 n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 3 n}

Take the partial fraction of

\frac{1}{n ^{2} + 3 n} : \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

= 3 \cdot n = 1 \sum \infty \frac{1}{3 n} - \frac{1}{3 ( n + 3 )}

Expand telescoping series:

\frac{11}{18}

Simplify

3 \cdot \frac{11}{18} : \frac{11}{6}

= \frac{11}{6}

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

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

\frac{d}{d x} (- x^{- x})

\frac{\partial}{\partial y} (cos (x) cos (y) + 2 x)

\frac{d}{d x} (2 y + x e^{y})

What is the sum from n=1 to infinity of 3/(n^2+3n) ?
The sum from n=1 to infinity of 3/(n^2+3n) is 11/6