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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{n !}{n ^{n}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

j = 2 \sum \infty j sin (\frac{10}{j})

k = 1 \sum \infty 2^{k}

n = 0 \sum \infty \frac{4}{8 ^{n}}

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

n = 0 \sum \infty \frac{2 ^{n}}{2 n - 1}

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