What is the sum from n=1 to infinity of 4/(n^2+5n+6) ?

The sum from n=1 to infinity of 4/(n^2+5n+6) is 4/3

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sum from n=1 to infinity of 4/(n^2+5n+6)

n = 1 \sum \infty \frac{4}{n ^{2} + 5 n + 6}

\frac{4}{3}

Solution steps

n = 1 \sum \infty \frac{4}{n ^{2} + 5 n + 6}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 5 n + 6}

Apply Telescoping Series Test:

\frac{1}{3}

= 4 \cdot \frac{1}{3}

Simplify

4 \cdot \frac{1}{3} : \frac{4}{3}

= \frac{4}{3}

\int \frac{sin ( \frac{9}{x} )}{x ^{2}} d x

\frac{\partial}{\partial x} (x^{2} - 10 x + y^{2} + 8)

\frac{1}{( s ^{3} + 4 s )}

10 (t - 1)

(2 x - 5 y + 2) d x + (1 - 6 y - 5 x) d y = 0

What is the sum from n=1 to infinity of 4/(n^2+5n+6) ?
The sum from n=1 to infinity of 4/(n^2+5n+6) is 4/3