What is the sum from n=0 to infinity of 2^n(1/2)^n ?

The sum from n=0 to infinity of 2^n(1/2)^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=0 to infinity of 2^n(1/2)^n

n = 0 \sum \infty 2^{n} (\frac{1}{2})^{n}

diverges

Solution steps

n = 0 \sum \infty 2^{n} (\frac{1}{2})^{n}

2^{n} (\frac{1}{2})^{n} = 1

= n = 0 \sum \infty 1

Every infinite sum of a non-zero constant diverges

= diverges

n = 1 \sum \infty \frac{n !}{( 4 n ) !}

n = 1 \sum \infty (\frac{7}{5})^{n}

n = 1 \sum \infty \frac{1}{12 n + 13}

n = 1 \sum \infty 5 n + 2

n = 0 \sum \infty \frac{3}{2} (\frac{1}{2})^{n}

What is the sum from n=0 to infinity of 2^n(1/2)^n ?
The sum from n=0 to infinity of 2^n(1/2)^n is diverges