What is the sum from n=0 to infinity of 1/(n^2+5n+4) ?

The sum from n=0 to infinity of 1/(n^2+5n+4) is converges

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sum from n=0 to infinity of 1/(n^2+5n+4)

n = 0 \sum \infty \frac{1}{n ^{2} + 5 n + 4}

converges

Solution steps

n = 0 \sum \infty \frac{1}{n ^{2} + 5 n + 4}

Apply Series Integral Test:

converges

= converges

n = 3 \sum \infty (\frac{2}{3})^{n}

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

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

m = 2 \sum \infty \frac{6}{7 ^{m}}

n = 1 \sum \infty \frac{1}{1 - e ^{- n}}

What is the sum from n=0 to infinity of 1/(n^2+5n+4) ?
The sum from n=0 to infinity of 1/(n^2+5n+4) is converges