What is the sum from n=1 to infinity of (64)/(n^3) ?

The sum from n=1 to infinity of (64)/(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 (64)/(n^3)

n = 1 \sum \infty \frac{64}{n ^{3}}

converges

Solution steps

n = 1 \sum \infty \frac{64}{n ^{3}}

Apply the constant multiplication rule:

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

= 64 \cdot n = 1 \sum \infty \frac{1}{n ^{3}}

Apply p-Series Test:

converges

= 64 converges

= converges

\int \frac{6 x ^{2} - 3 x + 1}{( 4 x + 1 ) ( x ^{2} + 1 )} d x

y = 2^{3 x^{2}}

\int \frac{( cos ^{5} ( t ) )}{sin ( t )} d t

x \to a lim (\frac{f ( x )}{g ( x )})

\frac{d}{d x} (2 cos (\frac{x}{2}) + π)

What is the sum from n=1 to infinity of (64)/(n^3) ?
The sum from n=1 to infinity of (64)/(n^3) is converges