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}

Simplify

(1 + \frac{6}{n})^{- n} : \frac{1}{\frac{( n + 6 ) ^{n}}{n ^{n}}}

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

Simplify

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

Apply Series Divergence Test:

diverges

= diverges

\frac{d}{d x} (\frac{4 x}{1 - tan ( x )})

\int \frac{1}{x ^{2} + 60} d x

x \to \infty lim (4 x e^{(\frac{1}{x})} - 4 x)

\int (2 y - \frac{1}{2 y}) d y

2 x^{2} - x^{4}

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