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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = \frac{π}{2} \sum \infty sin (n)

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

k = 1 \sum \infty \frac{9}{k ( k + 2 )}

n = 1 \sum \infty \frac{- 5}{n}

n = 1 \sum \infty \frac{n}{( 2 n + 1 ) ^{2}}

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