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

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

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

converges

Solution steps

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

Apply Series Integral Test:

converges

= converges

y^{^{'}} + 5 x y = 0

x \to - 1 lim (\frac{2}{x ^{2} - 1})

x^{3} - 7 x

\int sin (3 x y) d x

f (x) = x + x

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