What is the sum from n=0 to infinity of 1/(n^2+4n+3) ?

The sum from n=0 to infinity of 1/(n^2+4n+3) is converges

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sum from n=0 to infinity of 1/(n^2+4n+3)

n = 0 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

converges

Solution steps

n = 0 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

Apply Series Integral Test:

converges

= converges

y = 2 x^{3} - x^{2} + 5, (2, 17)

f (x) = 3 x^{2} + 4 x + 1, at x = 0

8 cos^{2} (y) d x + csc^{2} (x) d y = 0, y (\frac{π}{12}) = \frac{π}{4}

f (x) = \frac{x + 3}{x}

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

What is the sum from n=0 to infinity of 1/(n^2+4n+3) ?
The sum from n=0 to infinity of 1/(n^2+4n+3) is converges