What is the sum from n=0 to infinity of (n/(4-n))^3 ?

The sum from n=0 to infinity of (n/(4-n))^3 is diverges

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sum from n=0 to infinity of (n/(4-n))^3

n = 0 \sum \infty (\frac{n}{4 - n})^{3}

diverges

Solution steps

n = 5 \sum \infty (\frac{n}{4 - n})^{3}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{n}{9 ^{n}}

n = 0 \sum \infty \frac{2 n}{5 n - 3}

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

n = 3 \sum \infty \frac{n}{5 ^{n}}

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

What is the sum from n=0 to infinity of (n/(4-n))^3 ?
The sum from n=0 to infinity of (n/(4-n))^3 is diverges