What is the sum from n=0 to infinity of ln(n/(3n+1)) ?

The sum from n=0 to infinity of ln(n/(3n+1)) is diverges

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sum from n=0 to infinity of ln(n/(3n+1))

n = 0 \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

n = 1 \sum \infty \frac{5}{n ^{\frac{9}{8}}}

n = 0 \sum \infty sin (\frac{1}{n ^{0}})

n = 1 \sum \infty \frac{5}{n + 4}

n = 2 \sum \infty (- 0.95)^{n}

n = 1 \sum \infty 10 (- \frac{3}{8})^{n}

What is the sum from n=0 to infinity of ln(n/(3n+1)) ?
The sum from n=0 to infinity of ln(n/(3n+1)) is diverges