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 ln (\frac{1}{n !})

diverges

Solution steps

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

Apply log rule:

lo g_{a} (\frac{1}{x}) = - lo g_{a} (x)

= n = 1 \sum \infty - ln (n!)

Apply the constant multiplication rule:

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

= - n = 1 \sum \infty ln (n!)

Apply Series Divergence Test:

diverges

= - diverges

= diverges

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

n = 1 \sum \infty \frac{9 ^{n}}{n ^{5}}

n = 1 \sum \infty \frac{1}{9 n ^{3}}

n = 0 \sum \infty 5 (\frac{7}{8})^{n}

n = 0 \sum \infty \frac{1}{n ^{6} + 1}

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