What is the sum from n=2 to infinity of 4/(n^3) ?

The sum from n=2 to infinity of 4/(n^3) is converges

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sum from n=2 to infinity of 4/(n^3)

n = 2 \sum \infty \frac{4}{n ^{3}}

converges

Solution steps

n = 2 \sum \infty \frac{4}{n ^{3}}

Apply the constant multiplication rule:

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

= 4 \cdot n = 2 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 4 converges

= converges

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

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

n = 0 \sum \infty \frac{3}{8 ^{n}}

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

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

What is the sum from n=2 to infinity of 4/(n^3) ?
The sum from n=2 to infinity of 4/(n^3) is converges