What is the sum from n=0 to infinity of 2/(n^2+6n+8) ?

The sum from n=0 to infinity of 2/(n^2+6n+8) is converges

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sum from n=0 to infinity of 2/(n^2+6n+8)

n = 0 \sum \infty \frac{2}{n ^{2} + 6 n + 8}

converges

Solution steps

n = 0 \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 = 0 \sum \infty \frac{1}{n ^{2} + 6 n + 8}

Apply Series Integral Test:

converges

= 2 converges

= converges

(2 x - 5)^{2}

\int - \frac{16}{x ^{17}} d x

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

\frac{d y}{d x} = 2 y - 4 x e^{x}

x^{2} cos (x)

What is the sum from n=0 to infinity of 2/(n^2+6n+8) ?
The sum from n=0 to infinity of 2/(n^2+6n+8) is converges