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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= 3 converges

= converges

y = \frac{9}{x - 2}, (3, 9)

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

2 y^{^{''}} - 6 y^{^{'}} + y = 0

\int 5 sec^{2} (x) d x

6 x^{2} y^{^{'}} = y^{^{'}} + 6 x e^{- y}

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