What is the sum from n=1 to infinity of n/(e^{4n)} ?

The sum from n=1 to infinity of n/(e^{4n)} is converges

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sum from n=1 to infinity of n/(e^{4n)}

n = 1 \sum \infty \frac{n}{e ^{4 n}}

converges

Solution steps

n = 1 \sum \infty \frac{n}{e ^{4 n}}

Apply Series Ratio Test:

converges

= converges

\int \frac{2 x ^{2} - 5 x + 2}{x ^{3} + x} d x

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

\int (1 - 2 x)^{2021} d x

x \to + 4 + lim (ln (x - 4))

\frac{1}{( 1 - x ) ^{3}}, 0

What is the sum from n=1 to infinity of n/(e^{4n)} ?
The sum from n=1 to infinity of n/(e^{4n)} is converges