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

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

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

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

converges

Solution steps

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

Apply Series Comparison Test:

converges

= converges

- (3 x - 4)^{3}, 3

\int_{0}^{\frac{3 π}{2}} 5 ∣ sin (x) ∣ d x

\frac{x ^{2}}{2 + x}

\int_{- 1}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

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

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