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

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

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sum from n=1 to infinity of n*e^{-n^2}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{\partial}{\partial x} ((y + 1)^{2})

f (y) = 2 4 - y

x \to 9 lim (\frac{∣ 9 - x ∣}{9 - x})

x \to \infty lim (0. 2^{x})

\int (x - \frac{1}{2 x})^{2} d x

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