What is the sum from n=1 to infinity of 1/n (3/2)^n ?

The sum from n=1 to infinity of 1/n (3/2)^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 1/n (3/2)^n

n = 1 \sum \infty \frac{1}{n} (\frac{3}{2})^{n}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{n} (\frac{3}{2})^{n}

Simplify

\frac{1}{n} (\frac{3}{2})^{n} : \frac{3 ^{n}}{2 ^{n} n}

= n = 1 \sum \infty \frac{3 ^{n}}{2 ^{n} n}

Apply Series Ratio Test:

diverges

= diverges

x \to 4 lim (\frac{7 x - 1}{x - 4})

\frac{d}{d x} (3 e^{4 x^{2}})

\frac{s}{( s ^{2} + 9 ) \cdot ( s ^{2} + 16 )}

ln (\frac{1 + x}{1 - x})

\frac{\partial}{\partial y} (4 x^{3} y^{3} - 2)

What is the sum from n=1 to infinity of 1/n (3/2)^n ?
The sum from n=1 to infinity of 1/n (3/2)^n is diverges