What is the sum from n=0 to infinity of (2n)/(e^n) ?

The sum from n=0 to infinity of (2n)/(e^n) is converges

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sum from n=0 to infinity of (2n)/(e^n)

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

x \to 0 lim (\frac{( x ^{2} + 4 x )}{x})

(3 x^{2} + y - 4) - (2 y - x) y^{^{'}} = 0

2 y \cdot \frac{d y}{d x} = - x^{2}

\frac{d}{d x} (x k)

(2, - 3), (0, 4)

What is the sum from n=0 to infinity of (2n)/(e^n) ?
The sum from n=0 to infinity of (2n)/(e^n) is converges