What is the sum from n=1 to infinity of ne^{-4n^2} ?

The sum from n=1 to infinity of ne^{-4n^2} is converges

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sum from n=1 to infinity of ne^{-4n^2}

n = 1 \sum \infty n e^{- 4 n^{2}}

converges

Solution steps

n = 1 \sum \infty n e^{- 4 n^{2}}

Apply Series Ratio Test:

converges

= converges

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

n = 5 \sum \infty \frac{1}{ln ( n )}

n = 0 \sum \infty n^{- 1}

n = 0 \sum \infty 2 - 3 (0.8)^{n}

n = 1 \sum \infty \frac{6}{n ^{2} + n ^{3}}

What is the sum from n=1 to infinity of ne^{-4n^2} ?
The sum from n=1 to infinity of ne^{-4n^2} is converges