What is the sum from n=1 to infinity of n2^{-n} ?

The sum from n=1 to infinity of 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 n2^{-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\int x y e^{y^{2} x} d y

\frac{3}{e ^{x} + e ^{- x}}

\frac{d}{d x} (k e^{x})

x sin (\frac{y}{x}) \frac{d y}{d x} = y sin (\frac{y}{x}) + x

\int \frac{1}{a ^{2} + v ^{2}} d v

What is the sum from n=1 to infinity of n2^{-n} ?
The sum from n=1 to infinity of n2^{-n} is converges