What is the sum from n=1 to infinity of 6/(n^3) ?

The sum from n=1 to infinity of 6/(n^3) is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=1 to infinity of 6/(n^3)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply p-Series Test:

converges

= 6 converges

= converges

\int \frac{4 x ^{2} - 2 x}{x} d x

x \to 1 lim (\frac{x ^{3} - 1}{x - 1})

x \to 4 lim (3 (- 3) + 1)

\int (\frac{1}{5})^{x} d x

\frac{\partial}{\partial x} (arcsin (6 x yz))

What is the sum from n=1 to infinity of 6/(n^3) ?
The sum from n=1 to infinity of 6/(n^3) is converges