What is the sum from n=2 to infinity of ne^{-3n^2} ?

The sum from n=2 to infinity of ne^{-3n^2} is converges

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sum from n=2 to infinity of ne^{-3n^2}

n = 2 \sum \infty n e^{- 3 n^{2}}

converges

Solution steps

n = 2 \sum \infty n e^{- 3 n^{2}}

Apply Series Ratio Test:

converges

= converges

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

k = 3 \sum \infty \frac{1}{4 k ^{2} - 1}

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 1}

n = 2 \sum \infty \frac{n !}{( 2 n ) !}

n = 1 \sum \infty \frac{1}{n ^{3} + n}

What is the sum from n=2 to infinity of ne^{-3n^2} ?
The sum from n=2 to infinity of ne^{-3n^2} is converges