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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\int (e^{\frac{x}{2}}) d x

\frac{\partial}{\partial x} (2 x yz - y e^{2 x yz})

(3.1) (1.7)

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

\frac{d}{d x} (- sin (x^{2}) 2 x)

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