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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

- u

\frac{d y}{d x} y = 3 x^{3} - 8 x^{2} + π

\int \frac{csc ^{2} ( x )}{cot ^{5} ( x )} d x

n = 1 \sum \infty \frac{1}{1 3 ^{n} - 5 ^{n}}

(1, - 5), (- 5, 5)

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