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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 5 \sum \infty 2 (- \frac{1}{4})^{n}

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

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

n = 1 \sum \infty \frac{3 ^{n} + n}{n !}

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

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