What is the sum from n=0 to infinity of 1/(4+3n) ?

The sum from n=0 to infinity of 1/(4+3n) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of 1/(4+3n)

n = 0 \sum \infty \frac{1}{4 + 3 n}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{4 + 3 n}

Apply Series Limit Comparison Test:

diverges

= diverges

n = 2 \sum \infty \frac{3}{n ln ^{9} ( n )}

k = 2 \sum \infty \frac{1}{k ln ^{2} ( k )}

n = 0 \sum \infty \frac{n !}{50 0 ^{n}}

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

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

What is the sum from n=0 to infinity of 1/(4+3n) ?
The sum from n=0 to infinity of 1/(4+3n) is diverges