What is the sum from n=1 to infinity of 2/(nln(9n)) ?

The sum from n=1 to infinity of 2/(nln(9n)) is diverges

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sum from n=1 to infinity of 2/(nln(9n))

n = 1 \sum \infty \frac{2}{n ln ( 9 n )}

diverges

Solution steps

n = 1 \sum \infty \frac{2}{n ln ( 9 n )}

Apply the constant multiplication rule:

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

= 2 \cdot n = 1 \sum \infty \frac{1}{n ln ( 9 n )}

Apply Series Integral Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty \frac{n !}{11 6 ^{n}}

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

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

n = 0 \sum \infty \frac{1}{1 + n ^{4}}

n = 1 \sum \infty sin^{n} (7)

What is the sum from n=1 to infinity of 2/(nln(9n)) ?
The sum from n=1 to infinity of 2/(nln(9n)) is diverges