What is the sum from n=1 to infinity of (ln(6n))/n ?

The sum from n=1 to infinity of (ln(6n))/n is diverges

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sum from n=1 to infinity of (ln(6n))/n

n = 1 \sum \infty \frac{ln ( 6 n )}{n}

diverges

Solution steps

n = 1 \sum \infty \frac{ln ( 6 n )}{n}

Apply Series Integral Test:

diverges

= diverges

\frac{\frac{1}{2}}{s ^{2} + 2 s + 3}

3 (3 x^{2} + y^{2}) d x - 2 x y d y = 0

t + y e^{2 t y} + t e^{2 t y} \frac{d y}{d t} = 0

\frac{\partial}{\partial x} (2 e^{- 2 x y} + x y^{3})

\frac{d}{d x} (\frac{sin ( x )}{1 + tan ( x )})

What is the sum from n=1 to infinity of (ln(6n))/n ?
The sum from n=1 to infinity of (ln(6n))/n is diverges