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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Comparison Test:

converges

= 3 converges

= converges

n = 1 \sum \infty (0.5)^{n - 1}

n = 0 \sum \infty 2^{1 - n}

n = 0 \sum \infty 2 + (- 1)^{n}

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

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

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