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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty \frac{- 1}{n}

n = 1 \sum \infty \frac{6}{n ^{2} + 3}

n = 1 \sum \infty \frac{1}{n ^{\frac{9}{2}}}

n = 0 \sum \infty (- 0.4)^{n}

n = 1 \sum \infty 4 (- \frac{1}{5})^{4 n}

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