What is the sum from n=3 to infinity of 6/(n^3) ?

The sum from n=3 to infinity of 6/(n^3) is converges

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sum from n=3 to infinity of 6/(n^3)

n = 3 \sum \infty \frac{6}{n ^{3}}

converges

Solution steps

n = 3 \sum \infty \frac{6}{n ^{3}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 3 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 6 converges

= converges

n = 1 \sum \infty 4 (1.3)^{n}

n = 0 \sum \infty n^{^{''}} - n^{^{'}}

n = 0 \sum \infty (\frac{99}{100})^{n}

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

n = 1 \sum \infty (\frac{2}{10})^{n}

What is the sum from n=3 to infinity of 6/(n^3) ?
The sum from n=3 to infinity of 6/(n^3) is converges