What is the sum from n=1 to infinity of 8/(n^3) ?

The sum from n=1 to infinity of 8/(n^3) is converges

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sum from n=1 to infinity of 8/(n^3)

n = 1 \sum \infty \frac{8}{n ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{8}{n ^{3}}

Apply the constant multiplication rule:

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

= 8 \cdot n = 1 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 8 converges

= converges

\frac{d}{d x} (1 - 3 x^{2} + y^{2})

\int sec^{3} (π) x d x

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

y^{^{'}} = 3 y + 6

x \to 0 lim (\frac{x}{x - 2})

What is the sum from n=1 to infinity of 8/(n^3) ?
The sum from n=1 to infinity of 8/(n^3) is converges