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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

\frac{( s + 1 ) ^{2}}{( s + 2 ) ^{4}}

y = \frac{3}{x}, (4, \frac{3}{4})

\int \frac{( 2 x + 1 )}{x - 1} d x

y^{^{''}} + 3 y^{^{'}} - 18 y = 0

\frac{d}{d x} (\frac{2 sin ( x )}{3 + cos ( x )})

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