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

The sum from n=1 to infinity of ((n^3))/(3^{n^2)} is converges

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sum from n=1 to infinity of ((n^3))/(3^{n^2)}

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

\frac{d y}{d x} = e^{- 2 y}

y^{^{''}} - 4 y^{^{'}} + 7 y = t e^{t}

f (x) = \frac{2 x ln ( x ) + x ^{2} - 1}{x - 1}

x \to 0 + lim ((1 + x)^{\frac{5}{x}})

\frac{d y}{d t} = - 3 (y - 4)

What is the sum from n=1 to infinity of ((n^3))/(3^{n^2)} ?
The sum from n=1 to infinity of ((n^3))/(3^{n^2)} is converges