What is the sum from n=1 to infinity of 1/(3n^3+8) ?

The sum from n=1 to infinity of 1/(3n^3+8) is converges

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sum from n=1 to infinity of 1/(3n^3+8)

n = 1 \sum \infty \frac{1}{3 n ^{3} + 8}

converges

Solution steps

n = 1 \sum \infty \frac{1}{3 n ^{3} + 8}

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty e^{- \frac{n}{9}}

n = 1 \sum \infty \frac{1}{n ^{10}}

n = 1 \sum \infty cos (\frac{nπ}{2})

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

k = 2 \sum \infty \frac{1}{ln ( k )}

What is the sum from n=1 to infinity of 1/(3n^3+8) ?
The sum from n=1 to infinity of 1/(3n^3+8) is converges