What is the sum from n=1 to infinity of n/(e^{3n)} ?

The sum from n=1 to infinity of n/(e^{3n)} is converges

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sum from n=1 to infinity of n/(e^{3n)}

n = 1 \sum \infty \frac{n}{e ^{3 n}}

converges

Solution steps

n = 1 \sum \infty \frac{n}{e ^{3 n}}

Apply Series Ratio Test:

converges

= converges

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

\frac{d}{d x} (\frac{x ^{4}}{16} + \frac{1}{2 x ^{2}})

y = arccos (e^{3 x})

cos (3 x), 2 - cos (x)

f (x) = x^{3} - 3 x^{2} - 45 x + 5

What is the sum from n=1 to infinity of n/(e^{3n)} ?
The sum from n=1 to infinity of n/(e^{3n)} is converges