What is the sum from n=1 to infinity of (2/n)^3 ?

The sum from n=1 to infinity of (2/n)^3 is converges

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sum from n=1 to infinity of (2/n)^3

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

converges

Solution steps

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

Simplify

(\frac{2}{n})^{3} : \frac{8}{n ^{3}}

= 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

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

n = 0 \sum \infty (\frac{1}{π})^{n}

n = 0 \sum \infty 2 (\frac{3}{4})^{n}

n = 1 \sum \infty n e^{- 7 n^{2}}

k = 0 \sum \infty (0.6)^{k}

What is the sum from n=1 to infinity of (2/n)^3 ?
The sum from n=1 to infinity of (2/n)^3 is converges