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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 3 \sum \infty n^{2} e^{- n^{3}}

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

n = 1 \sum \infty \frac{n !}{7 ^{n}}

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

k = 1 \sum \infty k

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