What is the sum from n=1 to infinity of ln(1/(3^n)) ?

The sum from n=1 to infinity of ln(1/(3^n)) is diverges

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sum from n=1 to infinity of ln(1/(3^n))

n = 1 \sum \infty ln (\frac{1}{3 ^{n}})

diverges

Solution steps

n = 1 \sum \infty ln (\frac{1}{3 ^{n}})

Simplify

ln (\frac{1}{3 ^{n}}) : - ln (3) n

= n = 1 \sum \infty - ln (3) n

Apply the constant multiplication rule:

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

= - ln (3) \cdot n = 1 \sum \infty n

Apply Series Divergence Test:

diverges

= - ln (3) diverges

= diverges

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

n = 3 \sum \infty n e^{- n}

n = 1 \sum \infty \frac{5}{n ^{2} + 49}

n = 0 \sum \infty \frac{n}{n ^{2} - 2 n + 1}

n = 0 \sum \infty \frac{7}{n} - \frac{7}{n + 1}

What is the sum from n=1 to infinity of ln(1/(3^n)) ?
The sum from n=1 to infinity of ln(1/(3^n)) is diverges