What is the sum from n=1 to infinity of (ln(n^4))/n ?

The sum from n=1 to infinity of (ln(n^4))/n is diverges

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sum from n=1 to infinity of (ln(n^4))/n

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

n = 2 \sum \infty \frac{5}{n ( n + 2 )}

n = 1 \sum \infty \frac{1}{3 ^{n} - n}

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 2}

n = 0 \sum \infty e^{- 19 n}

n = 7 \sum \infty \frac{n !}{n - 6}

What is the sum from n=1 to infinity of (ln(n^4))/n ?
The sum from n=1 to infinity of (ln(n^4))/n is diverges