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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 6 \sum \infty \frac{2}{n - 5}

n = 0 \sum \infty \frac{1}{e ^{- n}}

n = 0 \sum \infty \frac{1}{1 + 3 n ^{3}}

n = 1 \sum \infty (\frac{5}{7})^{n}

n = 3 \sum \infty \frac{1}{2 ^{n} - 5}

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