What is the sum from n=1 to infinity of (n/(n+12))^n ?

The sum from n=1 to infinity of (n/(n+12))^n is diverges

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sum from n=1 to infinity of (n/(n+12))^n

n = 1 \sum \infty (\frac{n}{n + 12})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{n}{n + 12})^{n}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty n^{- 2}

n = 2 \sum \infty \frac{3}{5 ^{n - 1}}

n = 1 \sum \infty \frac{3}{1 + n ^{2}}

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

n = 1 \sum \infty cos (\frac{17}{n})

What is the sum from n=1 to infinity of (n/(n+12))^n ?
The sum from n=1 to infinity of (n/(n+12))^n is diverges