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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

Using Summation Formulas sum from n=1 to infinity of (-1/3)^(n-1)

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

x \to 0 lim (\frac{( sin ( cos ( x ) ) )}{( sec ( x ) )})

- \frac{d}{d t} (sin (2 t))

- \frac{d}{d x} y = 3 e^{x}

- \frac{d}{d t} (2 t)

- \frac{d}{d x} f (x) = e^{6 x}

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