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

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

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

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

diverges

Solution steps

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

Apply Series Root Test:

diverges

= diverges

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

n = 9 \sum \infty \frac{n !}{n ^{n}}

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

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

n = 1 \sum \infty 3 ln (\frac{n}{n + 1})

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