What is the sum from n=0 to infinity of 2ne^{-n^2} ?

The sum from n=0 to infinity of 2ne^{-n^2} is converges

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sum from n=0 to infinity of 2ne^{-n^2}

n = 0 \sum \infty 2 n e^{- n^{2}}

converges

Solution steps

n = 0 \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 = 0 \sum \infty n e^{- n^{2}}

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 1 \sum \infty (\frac{13}{2})^{n}

k = 1 \sum \infty \frac{4 k}{e ^{k}}

n = 0 \sum \infty (1 + \frac{1}{n})^{3 n}

n = 1 \sum \infty \frac{31}{n !}

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

What is the sum from n=0 to infinity of 2ne^{-n^2} ?
The sum from n=0 to infinity of 2ne^{-n^2} is converges