What is the sum from n=1 to infinity of 1/(3n+5) ?

The sum from n=1 to infinity of 1/(3n+5) 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=1 to infinity of 1/(3n+5)

n = 1 \sum \infty \frac{1}{3 n + 5}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{3 n + 5}

Apply Series Limit Comparison Test:

diverges

= diverges

\int \frac{1}{y - x} d x

n = 1 \sum \infty \frac{1}{n ^{2} + 5 n}

x \to - 4 lim (x^{2} - 16)

\frac{d}{d x} (\frac{x ^{\frac{1}{2}}}{3 x})

f (x) = x^{3} + 7 e^{x}

What is the sum from n=1 to infinity of 1/(3n+5) ?
The sum from n=1 to infinity of 1/(3n+5) is diverges