What is the sum from n=0 to infinity of (1+3/n)^n ?

The sum from n=0 to infinity of (1+3/n)^n is diverges

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sum from n=0 to infinity of (1+3/n)^n

n = 0 \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

n = 1 \sum \infty \frac{1}{5 ^{n} + 1}

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

n = 1 \sum \infty \frac{5}{n ^{2} + 6}

n = 100 \sum \infty \frac{1}{n}

n = 2 \sum \infty (- 0.4)^{n}

What is the sum from n=0 to infinity of (1+3/n)^n ?
The sum from n=0 to infinity of (1+3/n)^n is diverges