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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

(t - 1) cos (4 (t - 1))

(x - 1) y^{^{'}} + x y = 6 e^{x}

n = 1 \sum \infty \frac{n !}{n ^{n}}

\int tan^{3} (x) (sec (x))^{- \frac{1}{2}} d x

\int_{0}^{2} (2 + x) d x

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