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

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

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

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

n = 1 \sum \infty π

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

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

n = 1 \sum \infty n^{- 3}

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