What is the sum from n=1 to infinity of 4/(n^3) ?

The sum from n=1 to infinity of 4/(n^3) is converges

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sum from n=1 to infinity of 4/(n^3)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= 4 converges

= converges

y^{^{''}} + 36 y = - 2 tan (6 x)

\frac{d}{d x} ((\frac{sin ( x )}{cos ( 2 x )})^{3})

f^{^{'}} (x) = 3 cos (x) + 4 (- sin (x))

x \to 0 + lim (\frac{2}{ln ( x )})

\frac{d}{d x} (8 x^{2} ln (8 x))

What is the sum from n=1 to infinity of 4/(n^3) ?
The sum from n=1 to infinity of 4/(n^3) is converges