What is the sum from n=1 to infinity of 7/(n^3) ?

The sum from n=1 to infinity of 7/(n^3) is converges

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sum from n=1 to infinity of 7/(n^3)

n = 1 \sum \infty \frac{7}{n ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{7}{n ^{3}}

Apply the constant multiplication rule:

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

= 7 \cdot n = 1 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 7 converges

= converges

n = 1 \sum \infty 3 (\frac{4}{5})^{n}

n = 0 \sum \infty (- \frac{1}{9})^{n}

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

n = 0 \sum \infty \frac{4}{1 + 3 n ^{3}}

n = 0 \sum \infty \frac{n ^{2}}{5 + 2 n}

What is the sum from n=1 to infinity of 7/(n^3) ?
The sum from n=1 to infinity of 7/(n^3) is converges