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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

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

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

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

n = 0 \sum \infty \frac{n}{1 0 ^{n}}

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

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