What is the sum from n=1 to infinity of 1/(n^4+2n+8) ?

The sum from n=1 to infinity of 1/(n^4+2n+8) is converges

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sum from n=1 to infinity of 1/(n^4+2n+8)

n = 1 \sum \infty \frac{1}{n ^{4} + 2 n + 8}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{4} + 2 n + 8}

Apply Series Comparison Test:

converges

= converges

n = 0 \sum \infty \frac{n}{n + 4}

n = 0 \sum \infty \frac{3}{n ( n + 2 )}

n = 1 \sum \infty \frac{7}{n ^{2} + 64}

n = 2 \sum \infty \frac{- 2}{n ^{2} + n}

n = 1 \sum \infty \frac{1}{n ^{0.2}}

What is the sum from n=1 to infinity of 1/(n^4+2n+8) ?
The sum from n=1 to infinity of 1/(n^4+2n+8) is converges