What is the sum from n=0 to infinity of (1+2/n)^n ?

The sum from n=0 to infinity of (1+2/n)^n is diverges

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sum from n=0 to infinity of (1+2/n)^n

n = 0 \sum \infty (1 + \frac{2}{n})^{n}

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

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

n = 0 \sum \infty (- \frac{1}{7})^{n}

n = 1 \sum \infty \frac{1}{n ^{8}}

n = 1 \sum \infty \frac{n}{n ^{3} + 4}

What is the sum from n=0 to infinity of (1+2/n)^n ?
The sum from n=0 to infinity of (1+2/n)^n is diverges