What is the sum from n=2 to infinity of n^{ln(n)} ?

The sum from n=2 to infinity of n^{ln(n)} is diverges

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sum from n=2 to infinity of n^{ln(n)}

n = 2 \sum \infty n^{l n (n)}

diverges

Solution steps

n = 2 \sum \infty n^{l n (n)}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{8}{9 ^{n}}

n = 1 \sum \infty \frac{1}{n + 7}

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

n = 0 \sum \infty \frac{2}{n} - \frac{2}{n + 1}

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

What is the sum from n=2 to infinity of n^{ln(n)} ?
The sum from n=2 to infinity of n^{ln(n)} is diverges