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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Ratio Test:

converges

= 2 converges

= converges

(y^{3} - 2 x^{2} y) d x + 2 x^{3} d y = 0

\int x^{3} x^{2} + 7 d x

cos (θ)

\frac{\partial}{\partial x} (x ln (y - z))

x \to 2 lim (\frac{x ^{2} - 4}{x cos ( \frac{x π}{2} ) - 2 cos ( \frac{x π}{2} )})

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