What is the sum from n=2 to infinity of ne^{-5n^2} ?

The sum from n=2 to infinity of ne^{-5n^2} is converges

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sum from n=2 to infinity of ne^{-5n^2}

n = 2 \sum \infty n e^{- 5 n^{2}}

converges

Solution steps

n = 2 \sum \infty n e^{- 5 n^{2}}

Apply Series Ratio Test:

converges

= converges

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

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

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

n = 3 \sum \infty \frac{2}{3 ln ( n )}

n = 1 \sum \infty \frac{1}{9 n ^{4} + 3}

What is the sum from n=2 to infinity of ne^{-5n^2} ?
The sum from n=2 to infinity of ne^{-5n^2} is converges