What is the sum from n=1 to infinity of n*e^{-n} ?

The sum from n=1 to infinity of n*e^{-n} is converges

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sum from n=1 to infinity of n*e^{-n}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{5}{n + 1}

n = 0 \sum \infty \frac{2}{5 - n}

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

n = 0 \sum \infty sin (\frac{πn}{4})

n = 1 \sum \infty cosh (in)

What is the sum from n=1 to infinity of n*e^{-n} ?
The sum from n=1 to infinity of n*e^{-n} is converges