What is the sum from n=1 to infinity of 1/(n^2+4n+3) ?

The sum from n=1 to infinity of 1/(n^2+4n+3) is converges

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sum from n=1 to infinity of 1/(n^2+4n+3)

n = 1 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

Apply Series Integral Test:

converges

= converges

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\int_{- 2}^{3} \frac{32}{x ^{4}} d x

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- 2 sin (3 t)

x \to \frac{π}{2} lim (\frac{cos ( x )}{cot ( x )})

What is the sum from n=1 to infinity of 1/(n^2+4n+3) ?
The sum from n=1 to infinity of 1/(n^2+4n+3) is converges