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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Integral Test:

converges

= 4 converges

= converges

n = 5 \sum \infty cos (\frac{1}{n})

n = 1 \sum \infty n^{2} e^{- 3 n}

k = 1 \sum \infty \frac{1}{k ^{2} + 1}

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

n = 2 \sum \infty (- 0.7)^{n}

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