What is the sum from n=1 to infinity of (2^n)/n ?

The sum from n=1 to infinity of (2^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 (2^n)/n

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

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

\frac{d}{d x} (\frac{1}{2} (tan (x) - x))

n = 0 \sum \infty n^{3} x^{n}

\frac{d}{d x} (9 x arcsin (x))

\frac{d}{d x} (6 x - 5)

What is the sum from n=1 to infinity of (2^n)/n ?
The sum from n=1 to infinity of (2^n)/n is diverges