What is the sum from n=1 to infinity of (3^n)/(n2^n) ?

The sum from n=1 to infinity of (3^n)/(n2^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)/(n2^n)

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

\frac{\partial}{\partial y} (arctan (\frac{y}{x}))

y^{^{''}} + 18 y^{^{'}} + 97 y = 0

\frac{\partial}{\partial x} (arcsin (\frac{y}{( x ^{2} + y ^{2} ) ^{\frac{1}{2}}}))

\int tan (x) - 2 sin (x) d x

\frac{d}{d x} (\frac{300}{x})

What is the sum from n=1 to infinity of (3^n)/(n2^n) ?
The sum from n=1 to infinity of (3^n)/(n2^n) is diverges