What is the sum from n=1 to infinity of 3/(n^3+7) ?

The sum from n=1 to infinity of 3/(n^3+7) is converges

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sum from n=1 to infinity of 3/(n^3+7)

n = 1 \sum \infty \frac{3}{n ^{3} + 7}

converges

Solution steps

n = 1 \sum \infty \frac{3}{n ^{3} + 7}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ^{3} + 7}

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 1 \sum \infty (\frac{4}{7})^{n - 1}

n = 1 \sum \infty cos (\frac{8}{n})

n = 1 \sum \infty (cos (15))^{n}

n = 0 \sum \infty (\frac{π}{6})^{n}

n = 0 \sum \infty (- \frac{2}{9})^{n}

What is the sum from n=1 to infinity of 3/(n^3+7) ?
The sum from n=1 to infinity of 3/(n^3+7) is converges