What is the sum from n=0 to infinity of 4/(1+3n^3) ?

The sum from n=0 to infinity of 4/(1+3n^3) is converges

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sum from n=0 to infinity of 4/(1+3n^3)

n = 0 \sum \infty \frac{4}{1 + 3 n ^{3}}

converges

Solution steps

n = 0 \sum \infty \frac{4}{1 + 3 n ^{3}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 0 \sum \infty \frac{1}{1 + 3 n ^{3}}

Apply Series Comparison Test:

converges

= 4 converges

= converges

n = 0 \sum \infty \frac{n ^{2}}{5 + 2 n}

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

n = 1 \sum \infty \frac{1}{5 ^{n} - n}

i = 1 \sum \infty 5 \cdot 0. 2^{i - 1}

n = 1 \sum \infty \frac{3}{5 ^{n + 2}}

What is the sum from n=0 to infinity of 4/(1+3n^3) ?
The sum from n=0 to infinity of 4/(1+3n^3) is converges