What is the sum from n=0 to infinity of ((n+1)/n)^n ?

The sum from n=0 to infinity of ((n+1)/n)^n is diverges

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 ((n+1)/n)^n

n = 0 \sum \infty (\frac{n + 1}{n})^{n}

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

3 x^{^{''}} - 6 x^{^{'}} + 3 x = 0

x \to 0 + lim (\frac{5}{x} - \frac{5}{∣ x ∣})

y^{^{'}} = \frac{2 y ^{4} + x ^{4}}{x y ^{3}}

\frac{\partial}{\partial x} (2 k^{3} s^{2})

\int \frac{- 2}{x ^{2} - 1} d x

What is the sum from n=0 to infinity of ((n+1)/n)^n ?
The sum from n=0 to infinity of ((n+1)/n)^n is diverges