What is the sum from n=5 to infinity of 2(-1/4)^n ?

The sum from n=5 to infinity of 2(-1/4)^n is converges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=5 to infinity of 2(-1/4)^n

n = 5 \sum \infty 2 (- \frac{1}{4})^{n}

converges

Solution steps

n = 5 \sum \infty 2 (- \frac{1}{4})^{n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 5 \sum \infty (- \frac{1}{4})^{n}

Apply Series Ratio Test:

converges

= 2 converges

= converges

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

n = 1 \sum \infty \frac{1}{8 n + 4}

n = 1 \sum \infty \frac{3 ^{n} + n}{n !}

n = 0 \sum \infty \frac{n}{ln ( n )}

n = 1 \sum \infty \frac{e ^{n}}{n ^{3}}

What is the sum from n=5 to infinity of 2(-1/4)^n ?
The sum from n=5 to infinity of 2(-1/4)^n is converges