What is the sum from n=1 to infinity of (1-6/n)^n ?

The sum from n=1 to infinity of (1-6/n)^n is diverges

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sum from n=1 to infinity of (1-6/n)^n

n = 1 \sum \infty (1 - \frac{6}{n})^{n}

diverges

Solution steps

n = 1 \sum \infty (1 - \frac{6}{n})^{n}

Apply Series Divergence Test:

diverges

= diverges

k = 2 \sum \infty k e^{- 4 k^{2}}

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

n = 0 \sum \infty 0. 6^{n}

n = 0 \sum \infty cos (\frac{1}{n ^{3}})

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

What is the sum from n=1 to infinity of (1-6/n)^n ?
The sum from n=1 to infinity of (1-6/n)^n is diverges