What is the sum from n=1 to infinity of 6/(n^2+6n+8) ?

The sum from n=1 to infinity of 6/(n^2+6n+8) is converges

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sum from n=1 to infinity of 6/(n^2+6n+8)

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

converges

Solution steps

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

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} + 6 n + 8}

Apply Series Integral Test:

converges

= 6 converges

= converges

What is the sum from n=1 to infinity of 6/(n^2+6n+8) ?
The sum from n=1 to infinity of 6/(n^2+6n+8) is converges