What is the sum from k=0 to infinity of 2^k ?

The sum from k=0 to infinity of 2^k is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from k=0 to infinity of 2^k

k = 0 \sum \infty 2^{k}

diverges

Solution steps

k = 0 \sum \infty 2^{k}

Apply Series Geometric Test:

diverges

= diverges

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

k = 2 \sum \infty \frac{1}{k ^{2} - k}

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

k = 3 \sum \infty \frac{1}{k ^{2} - k}

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

What is the sum from k=0 to infinity of 2^k ?
The sum from k=0 to infinity of 2^k is diverges