What is the sum from n=1 to infinity of 5/(n^2+4n+3) ?

The sum from n=1 to infinity of 5/(n^2+4n+3) is converges

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sum from n=1 to infinity of 5/(n^2+4n+3)

n = 1 \sum \infty \frac{5}{n ^{2} + 4 n + 3}

converges

Solution steps

n = 1 \sum \infty \frac{5}{n ^{2} + 4 n + 3}

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty \frac{1}{n ^{2} + 4 n + 3}

Apply Series Integral Test:

converges

= 5 converges

= converges

n = 0 \sum \infty (e)^{- n^{2}}

n = 1 \sum \infty cos (\frac{13}{n})

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

n = 1 \sum \infty (\frac{12}{7})^{n}

n = 1 \sum \infty n (\frac{5}{6})^{n - 1}

What is the sum from n=1 to infinity of 5/(n^2+4n+3) ?
The sum from n=1 to infinity of 5/(n^2+4n+3) is converges