What is the sum from k=1 to infinity of ke^{-3k} ?

The sum from k=1 to infinity of ke^{-3k} is converges

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sum from k=1 to infinity of ke^{-3k}

k = 1 \sum \infty k e^{- 3 k}

converges

Solution steps

k = 1 \sum \infty k e^{- 3 k}

Apply Series Ratio Test:

converges

= converges

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

n = 1 \sum \infty \frac{1}{( n + 3 ) !}

n = 0 \sum \infty \frac{1}{n ln ^{2} ( n )}

n = 1 \sum \infty 70 (0.1)^{n}

n = 1 \sum \infty \frac{8 ^{n}}{7 ^{n}}

What is the sum from k=1 to infinity of ke^{-3k} ?
The sum from k=1 to infinity of ke^{-3k} is converges