What is the sum from n=0 to infinity of 2e^{-n} ?

The sum from n=0 to infinity of 2e^{-n} is 2*e/(e-1)

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sum from n=0 to infinity of 2e^{-n}

n = 0 \sum \infty 2 e^{- n}

2 \cdot \frac{e}{e - 1}

Solution steps

n = 0 \sum \infty 2 e^{- n}

Apply the constant multiplication rule:

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

= 2 \cdot n = 0 \sum \infty e^{- n}

Simplify

e^{- n} : \frac{1}{e ^{n}}

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

Apply Series Geometric Test:

\frac{e}{e - 1}

= 2 \cdot \frac{e}{e - 1}

\frac{\partial}{\partial x} (9 - x^{2})

\frac{d}{d x} (\frac{( 3 x ^{2} )}{2 x + 4})^{3}

x \to 0 lim (\frac{3}{x} + \frac{3}{x ^{2} - x})

\frac{d}{d x} (2 x + sin (x))

\frac{d}{d x} (\frac{x ^{2} cos ( x )}{tan ( x )})

What is the sum from n=0 to infinity of 2e^{-n} ?
The sum from n=0 to infinity of 2e^{-n} is 2*e/(e-1)