What is the sum from n=2 to infinity of 1/(4n^3) ?

The sum from n=2 to infinity of 1/(4n^3) is converges

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sum from n=2 to infinity of 1/(4n^3)

n = 2 \sum \infty \frac{1}{4 n ^{3}}

converges

Solution steps

n = 2 \sum \infty \frac{1}{4 n ^{3}}

Apply the constant multiplication rule:

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

= \frac{1}{4} \cdot n = 2 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= \frac{1}{4} converges

= converges

n = 2 \sum \infty \frac{2 ^{n}}{n !}

n = 1 \sum \infty \frac{2 ^{n}}{n 2 ^{n}}

n = 1 \sum \infty 0. 5^{n - 1}

n = 1 \sum \infty π^{\frac{1}{n}}

n = 0 \sum \infty \frac{4 ^{n}}{n}

What is the sum from n=2 to infinity of 1/(4n^3) ?
The sum from n=2 to infinity of 1/(4n^3) is converges