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

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

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

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

converges

Solution steps

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

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} + 4 n + 3}

Apply Series Integral Test:

converges

= 2 converges

= converges

\int x - x^{2} d x

\frac{d}{d x} (\frac{- 4 x ^{6} + 4 x ^{5} - 3 x ^{3}}{3 x ^{2}})

\int e^{2 s i n^{2} (x)} sin (2 x) d x

\frac{50}{x}

\int \frac{20}{1 - cos ( 4 x )} d x

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