What is the sum from k=1 to infinity of ke^{-3k^2} ?

The sum from k=1 to infinity of ke^{-3k^2} is converges

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sum from k=1 to infinity of ke^{-3k^2}

k = 1 \sum \infty k e^{- 3 k^{2}}

converges

Solution steps

k = 1 \sum \infty k e^{- 3 k^{2}}

Apply Series Ratio Test:

converges

= converges

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

n = 2 \sum \infty \frac{3 ^{n}}{n !}

n = 1 \sum \infty \frac{8}{n + 2} - \frac{8}{n}

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

n = 1 \sum \infty 6^{- n}

What is the sum from k=1 to infinity of ke^{-3k^2} ?
The sum from k=1 to infinity of ke^{-3k^2} is converges