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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{n !}{n \cdot 2 ^{n}}

n = 1 \sum \infty \frac{1}{25 n ^{2} + 5 n - 6}

n = 0 \sum \infty (- 1)^{n} \frac{1}{n}

n = 1 \sum \infty 2 (- \frac{4}{3})^{n}

n = 2 \sum \infty \frac{1}{n ^{1} ln ( n )}

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