What is the sum from n=0 to infinity of 1/(ln(6n)) ?

The sum from n=0 to infinity of 1/(ln(6n)) is diverges

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sum from n=0 to infinity of 1/(ln(6n))

n = 0 \sum \infty \frac{1}{ln ( 6 n )}

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{12 n + 1}

k = 0 \sum \infty \frac{1}{k}

n = 1 \sum \infty (- \frac{6}{25})^{n}

n = 0 \sum \infty \frac{4}{3 n + 2}

n = 1 \sum \infty \frac{1}{12 n + 7}

What is the sum from n=0 to infinity of 1/(ln(6n)) ?
The sum from n=0 to infinity of 1/(ln(6n)) is diverges