What is the sum from n=1 to infinity of 6ne^{-n^2} ?

The sum from n=1 to infinity of 6ne^{-n^2} is converges

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sum from n=1 to infinity of 6ne^{-n^2}

n = 1 \sum \infty 6 n e^{- n^{2}}

converges

Solution steps

n = 1 \sum \infty 6 n e^{- n^{2}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty n e^{- n^{2}}

Apply Series Ratio Test:

converges

= 6 converges

= converges

k = 1 \sum \infty \frac{8}{k ^{7} + 9}

n = 0 \sum \infty \frac{2 n}{5 n + 1}

n = 1 \sum \infty n e^{- \frac{n}{9}}

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

n = 1 \sum \infty \frac{6}{4 n ^{2} + 4 n}

What is the sum from n=1 to infinity of 6ne^{-n^2} ?
The sum from n=1 to infinity of 6ne^{-n^2} is converges