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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= \frac{1}{9} converges

= converges

n = 0 \sum \infty 5 (\frac{7}{8})^{n}

n = 0 \sum \infty \frac{1}{n ^{6} + 1}

n = 0 \sum \infty \frac{3 n}{n !}

n = 1 \sum \infty \frac{1}{( n ln ( n ) )}

n = 0 \sum \infty \frac{i}{n ( n + 1 )}

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