What is the sum from n=0 to infinity of 1/(ln(n^n)) ?

The sum from n=0 to infinity of 1/(ln(n^n)) is diverges

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sum from n=0 to infinity of 1/(ln(n^n))

n = 0 \sum \infty \frac{1}{ln ( n ^{n} )}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{ln ( n ^{n} )}

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{4 + n ^{2}}

n = 1 \sum \infty \frac{ln ( 7 n )}{n}

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

n = 0 \sum \infty 2^{n} (\frac{1}{2})^{n}

n = 1 \sum \infty \frac{n !}{( 4 n ) !}

What is the sum from n=0 to infinity of 1/(ln(n^n)) ?
The sum from n=0 to infinity of 1/(ln(n^n)) is diverges