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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d ^{6}}{d x ^{6}} (x^{- 6})

\frac{\partial}{\partial x} ((x^{2} - a - 1) (x^{2} + a - 1))

\frac{\partial}{\partial x} (3 x^{2} + y^{2})

y = 3 x^{2} + e^{2} x + π^{2}

y = (1 + 4 (2))^{\frac{3}{2}}

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