What is the sum from n=1 to infinity of 1/(3(n^3+4)) ?

The sum from n=1 to infinity of 1/(3(n^3+4)) is converges

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sum from n=1 to infinity of 1/(3(n^3+4))

n = 1 \sum \infty \frac{1}{3 ( n ^{3} + 4 )}

converges

Solution steps

n = 1 \sum \infty \frac{1}{3 ( n ^{3} + 4 )}

Apply the constant multiplication rule:

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

= \frac{1}{3} \cdot n = 1 \sum \infty \frac{1}{n ^{3} + 4}

Apply Series Comparison Test:

converges

= \frac{1}{3} converges

= converges

n = 1 \sum \infty n^{2} (\frac{1}{3})^{n}

\int 1 - \frac{x}{y - x ^{2}}

\frac{d y}{d x} = \frac{1}{10} \cdot y \cdot (500 - y)

\frac{d}{d x} (sin^{2} (x + 5))

\frac{d}{d x} ((e^{3})^{x - 1})

What is the sum from n=1 to infinity of 1/(3(n^3+4)) ?
The sum from n=1 to infinity of 1/(3(n^3+4)) is converges