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

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

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

n = 2 \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

k = 1 \sum \infty (cos (10))^{k}

n = 1 \sum \infty \frac{( - 1 ) ^{n}}{3}

n = 1 \sum \infty cos (\frac{16}{n})

n = 1 \sum \infty \frac{1}{n ^{4} + 2 n + 8}

n = 0 \sum \infty \frac{n}{n + 4}

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