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

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

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

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

diverges

Solution steps

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

(\frac{n - 2}{n}) = (\frac{n}{n} - \frac{2}{n})

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

Simplify

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

Apply Series Divergence Test:

diverges

= diverges

\int (sin (2 u) - cos (3 u))^{(2)} d u

\frac{d}{d x} (- 3 x - 9)

\int \frac{y}{( 1 + y ^{2} ) ^{4}} d y

[x + (x + sin^{2} (x))^{3}]^{4}

x \to 2 + lim (\frac{x ^{2} - 4}{x - 2})

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