What is the sum from n=0 to infinity of (2-4/n)^5 ?

The sum from n=0 to infinity of (2-4/n)^5 is diverges

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sum from n=0 to infinity of (2-4/n)^5

n = 0 \sum \infty (2 - \frac{4}{n})^{5}

diverges

Solution steps

n = 1 \sum \infty (2 - \frac{4}{n})^{5}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{9}{n ^{2} + 64}

n = 6 \sum \infty \frac{1}{n} - \frac{1}{n + 3}

n = 0 \sum \infty \frac{1}{8 n - 1}

n = 1 \sum \infty (sin (200))^{n}

What is the sum from n=0 to infinity of (2-4/n)^5 ?
The sum from n=0 to infinity of (2-4/n)^5 is diverges