What is the sum from k=1 to infinity of ke^{-4k} ?

The sum from k=1 to infinity of ke^{-4k} is converges

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sum from k=1 to infinity of ke^{-4k}

k = 1 \sum \infty k e^{- 4 k}

converges

Solution steps

k = 1 \sum \infty k e^{- 4 k}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty 10 ln (8 - n)

n = 1 \sum \infty (1 + \frac{6}{n})^{n}

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

n = 5 \sum \infty \frac{1 ^{n}}{n !}

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

What is the sum from k=1 to infinity of ke^{-4k} ?
The sum from k=1 to infinity of ke^{-4k} is converges