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

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

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

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

diverges

Solution steps

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

(\frac{n + 2}{n}) = (\frac{n}{n} + \frac{2}{n})

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

Simplify

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

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{7}{n ^{\frac{2}{5}}}

n = 1 \sum \infty \frac{6 n}{3 n + 9}

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

n = 0 \sum \infty cos (\frac{nπ}{3})

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

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