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

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

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

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

\frac{3}{2}

Solution steps

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

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 + 2}

Apply Telescoping Series Test:

\frac{1}{2}

= 3 \cdot \frac{1}{2}

Simplify

3 \cdot \frac{1}{2} : \frac{3}{2}

= \frac{3}{2}

n = 1 \sum \infty \frac{2}{3 n + 8}

n = 1 \sum \infty \frac{1}{11 n ^{3} + 5}

k = 1 \sum \infty \frac{1}{2 + sin ( k )}

n = 1 \sum \infty \frac{6}{n ^{2} + 6 n + 8}

n = 0 \sum \infty \frac{n}{5 n ^{3} + 1}

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