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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 0 \sum \infty 2^{n} + i

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

n = 1 \sum \infty \frac{3}{9 ^{n + 1}}

n = 1 \sum \infty 2 n + 3

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

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