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

The sum from n=1 to infinity of (n^4)/(4n!) 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=1 to infinity of (n^4)/(4n!)

n = 1 \sum \infty \frac{n ^{4}}{4 n !}

converges

Solution steps

n = 1 \sum \infty \frac{n ^{4}}{4 n !}

Apply the constant multiplication rule:

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

= \frac{1}{4} \cdot n = 1 \sum \infty \frac{n ^{4}}{n !}

Simplify

n^{4} : n^{3} n

= \frac{1}{4} \cdot n = 1 \sum \infty \frac{n ^{3} n}{n !}

Apply Series Ratio Test:

converges

= \frac{1}{4} converges

= converges

x = 1 \sum \infty \frac{x}{x ( x + 1 )}

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

k = 1 \sum \infty k e^{- 16 k}

k = 1 \sum \infty k e^{- 15 k}

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

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