What is the sum from n=1 to infinity of 2/(1+e^n) ?

The sum from n=1 to infinity of 2/(1+e^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 2/(1+e^n)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

\frac{d}{d x} (\frac{( x - 3 )}{x + 3})

\int_{0}^{4} (e^{- 0.25 x} + 5) d x

\frac{1}{( s ^{2} + 1 ) ( s ^{2} )}

x \to 0 + lim (arccot (x))

\int (2 x^{2} + 3 x - 3) (5 x + 3) d x

What is the sum from n=1 to infinity of 2/(1+e^n) ?
The sum from n=1 to infinity of 2/(1+e^n) is converges