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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Integral Test:

converges

= 3 converges

= converges

x \to - 4 + lim (\frac{x + 5}{x + 4})

2 x^{3} - x^{2} - 5 x, - x^{2} + 3 x, 0, 2

\frac{d y}{d x} = \frac{1}{4 - x ^{2}}, y (0) = π

\int 5 tan^{3} (x) sec (x) d x

y^{^{'}} = (1 + t) \cdot y, y (0) = 1

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