What is the sum from n=1 to infinity of 6n^2e^{-n^3} ?

The sum from n=1 to infinity of 6n^2e^{-n^3} is converges

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sum from n=1 to infinity of 6n^2e^{-n^3}

n = 1 \sum \infty 6 n^{2} e^{- n^{3}}

converges

Solution steps

n = 1 \sum \infty 6 n^{2} e^{- n^{3}}

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty n^{2} e^{- n^{3}}

Apply Series Ratio Test:

converges

= 6 converges

= converges

n = 1 \sum \infty \frac{1}{n ^{\frac{5}{4}}}

n = 1 \sum \infty \frac{7}{n} - \frac{7}{n ^{2}}

n = 1 \sum \infty n e^{- 5 n^{2}}

n = 1 \sum \infty (0.8)^{n}

n = 1 \sum \infty (\frac{3}{7})^{n}

What is the sum from n=1 to infinity of 6n^2e^{-n^3} ?
The sum from n=1 to infinity of 6n^2e^{-n^3} is converges