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

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

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

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

converges

Solution steps

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

Simplify

e^{- 3 n} : \frac{1}{e ^{3 n}}

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

Simplify

n^{2} \frac{1}{e ^{3 n}} : \frac{n ^{2}}{e ^{3 n}}

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

Apply Series Ratio Test:

converges

= converges

k = 1 \sum \infty \frac{1}{k ^{2} + 1}

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

n = 2 \sum \infty (- 0.7)^{n}

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

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

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