What is the sum from n=3 to infinity of (6/11)^{-n} ?

The sum from n=3 to infinity of (6/11)^{-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=3 to infinity of (6/11)^{-n}

n = 3 \sum \infty (\frac{6}{11})^{- n}

diverges

Solution steps

n = 3 \sum \infty (\frac{6}{11})^{- n}

Apply Series Ratio Test:

diverges

= diverges

4 x (x^{2} - 9)

\frac{\partial}{\partial y} (x (1 - x - 2 y))

(0, 8), (8, 1)

x, at x = 16

\int_{2}^{x} (3 t^{2} - 1) d t

What is the sum from n=3 to infinity of (6/11)^{-n} ?
The sum from n=3 to infinity of (6/11)^{-n} is diverges