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

The sum from n=1 to infinity of 1/(3n+1) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

n = 1 \sum \infty \frac{1}{3 n + 1}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{3 n + 1}

Apply Series Integral Test:

diverges

= diverges

\frac{d}{d x} (csc (x^{2}))

\int sin (- 5 x^{4} - x - 6) (- 20 x^{3} - 1) d x

a re a f (x) = x^{2}, g (x) = 9 + x

\int \frac{y}{y ^{2} + 2} d y

\int (10 x - 3) d x

What is the sum from n=1 to infinity of 1/(3n+1) ?
The sum from n=1 to infinity of 1/(3n+1) is diverges