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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d}{d x} (\frac{1}{2 x + sin ^{3} ( x )})

\frac{d}{d x} ((2 x^{2} - 6 x + 1)^{- 8})

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

y^{^{''}} + 16 y = 8 tan (4 x)

\frac{2 y ^{^{'}}}{y} = x

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