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

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

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

converges

Solution steps

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

Simplify

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

Apply Series Root Test:

converges

= converges

n = 1 \sum \infty \frac{3}{n ^{2} + 3 n + 2}

n = 1 \sum \infty \frac{2}{3 n + 8}

n = 1 \sum \infty \frac{1}{11 n ^{3} + 5}

k = 1 \sum \infty \frac{1}{2 + sin ( k )}

n = 1 \sum \infty \frac{6}{n ^{2} + 6 n + 8}

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