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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\int \frac{3}{x ^{- 1} + 3} d x

2 x^{2} + x + 1

\frac{d}{d x} (2 x - \frac{2}{x})

\frac{d y}{d t} = - 0.0077 y

y^{^{'}} = 0.1 - \frac{y}{400}

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