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}

Decimal

1.33333 \dots

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}

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