What is the sum from k=5 to infinity of 3(1/3)^{3k} ?

The sum from k=5 to infinity of 3(1/3)^{3k} is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

sum from k=5 to infinity of 3(1/3)^{3k}

k = 5 \sum \infty 3 (\frac{1}{3})^{3 k}

converges

Solution steps

k = 5 \sum \infty 3 (\frac{1}{3})^{3 k}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 3 \cdot k = 5 \sum \infty (\frac{1}{3})^{3 k}

Apply Series Ratio Test:

converges

= 3 converges

= converges

What is the sum from k=5 to infinity of 3(1/3)^{3k} ?
The sum from k=5 to infinity of 3(1/3)^{3k} is converges