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

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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

x \to 0 lim (\frac{( cos ( x ) )}{x})

\frac{d}{d x} (\frac{1}{1 + 4 x ^{2}})

\frac{d}{d t} (π t)

x \to 0 lim (\frac{( sin ( 8 x ) )}{x})

x \to 2 lim (8 x^{3} + 1)

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