What is the sum from n=1 to infinity of 9/(n^3) ?

The sum from n=1 to infinity of 9/(n^3) is converges

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sum from n=1 to infinity of 9/(n^3)

n = 1 \sum \infty \frac{9}{n ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{9}{n ^{3}}

Apply the constant multiplication rule:

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

= 9 \cdot n = 1 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 9 converges

= converges

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

\frac{d}{d x} (\frac{ln ( x + 2 )}{4 x ^{- 2}})

\frac{2 s + 1}{s ^{3}}

h (x) = \frac{x ^{2} + 4 x + 1}{3 x + 5}

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

What is the sum from n=1 to infinity of 9/(n^3) ?
The sum from n=1 to infinity of 9/(n^3) is converges