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

\int 6 - x^{2} d x

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

2 x^{3} + 3 x^{2} - 12 x + 7

\frac{d}{d t} (\frac{t ^{3}}{3})

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