What is the sum from n=1 to infinity of 3/(nln(9n)) ?

The sum from n=1 to infinity of 3/(nln(9n)) is diverges

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sum from n=1 to infinity of 3/(nln(9n))

n = 1 \sum \infty \frac{3}{n ln ( 9 n )}

diverges

Solution steps

n = 1 \sum \infty \frac{3}{n ln ( 9 n )}

Apply the constant multiplication rule:

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

= 3 \cdot n = 1 \sum \infty \frac{1}{n ln ( 9 n )}

Apply Series Integral Test:

diverges

= 3 diverges

= diverges

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What is the sum from n=1 to infinity of 3/(nln(9n)) ?
The sum from n=1 to infinity of 3/(nln(9n)) is diverges