What is the sum from n=1 to infinity of (1+6/n)^n ?

The sum from n=1 to infinity of (1+6/n)^n is diverges

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sum from n=1 to infinity of (1+6/n)^n

n = 1 \sum \infty (1 + \frac{6}{n})^{n}

diverges

Solution steps

n = 1 \sum \infty (1 + \frac{6}{n})^{n}

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{n ^{2}}{2 n}

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

n = 1 \sum \infty \frac{- 1 ^{n}}{e ^{n}}

n = 1 \sum \infty (\frac{6}{9})^{n - 1}

n = 0 \sum \infty 3 (1 - n)^{- 2}

What is the sum from n=1 to infinity of (1+6/n)^n ?
The sum from n=1 to infinity of (1+6/n)^n is diverges