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

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

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

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 0 \sum \infty (- 3)^{- n}

n = 1 \sum \infty \frac{n !}{8 4 ^{n}}

n = 2 \sum \infty \frac{4}{ln ( n ^{2} )}

n = 0 \sum \infty \frac{1}{1 + 8 n}

k = 1 \sum \infty \frac{1}{8 ^{k} + 4}

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