What is the sum from n=1 to infinity of 1/(n^3+6) ?

The sum from n=1 to infinity of 1/(n^3+6) is converges

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sum from n=1 to infinity of 1/(n^3+6)

n = 1 \sum \infty \frac{1}{n ^{3} + 6}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{3} + 6}

Apply Series Comparison Test:

converges

= converges

n = 1 \sum \infty \frac{1}{n \cdot 2 ^{n}}

n = 0 \sum \infty \frac{- 4}{e ^{n}}

n = 1 \sum \infty (\frac{π}{e})^{- n}

n = 3 \sum \infty n

n = 1 \sum \infty cos (7 n)

What is the sum from n=1 to infinity of 1/(n^3+6) ?
The sum from n=1 to infinity of 1/(n^3+6) is converges