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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty 2^{n} 3^{n - 2}

n = 5 \sum \infty \frac{ln ( n )}{n}

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

n = 3 \sum \infty 4 (\frac{1}{2})^{n + 2}

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

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