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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\frac{d}{d x} (ln (1 + 9 x))

e^{- x} y \frac{d y}{d x} = e^{- y} + e^{- 2 x - y}

\int \frac{1}{x ^{2} - 9} d x

\int \frac{4 e ^{3 x}}{1 + e ^{3 x}} d x

x \to 0 lim (1 + 3^{x})

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