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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{6}{n ^{3}}

k = 0 \sum \infty (\frac{1}{8})^{k}

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

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

n = 4 \sum \infty \frac{3 ^{n}}{n ^{3}}

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