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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{1 + 8 n}

n = 0 \sum \infty \frac{3}{n}

n = 1 \sum \infty 4 - \frac{1}{n}

n = 0 \sum \infty \frac{7}{1 0 ^{- n}}

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

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