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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply log rule:

lo g_{a} (1) = 0

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

Apply Cauchy's Convergence Condition:

diverges

= 0 diverges

= diverges

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

k = 2 \sum \infty (- 0.8)^{k}

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

n = 2 \sum \infty \frac{n ^{e}}{n ^{π}}

n = 1 \sum \infty - 4 (\frac{5}{8})^{n}

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