What is the sum from n=0 to infinity}(2^{2n of)/(n!) ?

The sum from n=0 to infinity}(2^{2n of)/(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=0 to infinity}(2^{2n of)/(n!)

n = 0 \sum \infty \frac{2 ^{2 n}}{n !}

converges

Solution steps

n = 0 \sum \infty \frac{2 ^{2 n}}{n !}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty e^{0 n}

n = 1 \sum \infty n! e^{- 8 n}

n = 1 \sum \infty \frac{1}{n + sin ( n )}

n = 3 \sum \infty \frac{4}{ln ( n )}

n = 0 \sum \infty \frac{1}{7}

What is the sum from n=0 to infinity}(2^{2n of)/(n!) ?
The sum from n=0 to infinity}(2^{2n of)/(n!) is converges