What is the sum from n=1 to infinity of 3ne^{-n^2} ?

The sum from n=1 to infinity of 3ne^{-n^2} is converges

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sum from n=1 to infinity of 3ne^{-n^2}

n = 1 \sum \infty 3 n e^{- n^{2}}

converges

Solution steps

n = 1 \sum \infty 3 n e^{- n^{2}}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty n e^{- n^{2}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\int (1 + 9 - x^{2}) d x

\frac{d}{d x} (4 x^{3} - 6 x)

3 - e^{2 t}

f (x) = 2 - x

f (x) = \frac{5 x + 1}{x - 1}

What is the sum from n=1 to infinity of 3ne^{-n^2} ?
The sum from n=1 to infinity of 3ne^{-n^2} is converges