What is the sum from n=1 to infinity of ne^{-6n^2} ?

The sum from n=1 to infinity of ne^{-6n^2} is converges

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sum from n=1 to infinity of ne^{-6n^2}

n = 1 \sum \infty n e^{- 6 n^{2}}

converges

Solution steps

n = 1 \sum \infty n e^{- 6 n^{2}}

Apply Series Ratio Test:

converges

= converges

n = 1 \sum \infty \frac{( 4 - 3 ) ^{n}}{n}

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

n = 1 \sum \infty sin^{n} (8)

n = 1 \sum \infty \frac{1}{5 ^{n} + 3}

n = 1 \sum \infty \frac{n !}{10 4 ^{n}}

What is the sum from n=1 to infinity of ne^{-6n^2} ?
The sum from n=1 to infinity of ne^{-6n^2} is converges