What is the sum from n=3 to infinity of n/(e^{3n)} ?

The sum from n=3 to infinity of n/(e^{3n)} is converges

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sum from n=3 to infinity of n/(e^{3n)}

n = 3 \sum \infty \frac{n}{e ^{3 n}}

converges

Solution steps

n = 3 \sum \infty \frac{n}{e ^{3 n}}

Apply Series Ratio Test:

converges

= converges

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

x \to 2 lim (6)

x \to 0 lim (\frac{sin ( x ) - x}{x})

n = 1 \sum \infty 4 (\frac{3}{4})^{n}

\int 2 x \cdot cos (1 + 3 x^{2}) d x

What is the sum from n=3 to infinity of n/(e^{3n)} ?
The sum from n=3 to infinity of n/(e^{3n)} is converges