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

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

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

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

converges

Solution steps

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

Simplify

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

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

Simplify

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

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

Apply Series Ratio Test:

converges

= converges

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

n = 0 \sum \infty (1 - n)^{- 2}

n = 0 \sum \infty \frac{3}{2 n}

n = 0 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

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

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