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

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

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/(n2^n)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d y}{d x} = 12 x^{11} y

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

\int_{- 1}^{1} 1.5 x^{4} d x

\int \frac{x ^{2} + 2}{x + 1} d x

\int \frac{( 2 x )}{( 9 - x ^{4} )} d x

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