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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 2 converges

= converges

\frac{d}{d x} (\frac{x - 2}{x ^{2} - 4})

y = x^{2} + x + 9, at x = 2

\frac{d}{d x} ((\frac{x ^{2} + 2}{x ^{2} - 2})^{8})

(- 4, 2), (4, - 2)

f (x) = \frac{x ^{3}}{2}, at x = - 6

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