What is the sum from n=0 to infinity of-ln(3-n) ?

The sum from n=0 to infinity of-ln(3-n) is diverges

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sum from n=0 to infinity of-ln(3-n)

n = 0 \sum \infty - ln (3 - n)

diverges

Solution steps

n = 4 \sum \infty - ln (3 - n)

Apply the constant multiplication rule:

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

= - n = 4 \sum \infty ln (3 - n)

Apply Series Divergence Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty (0.6)^{n}

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

n = 2 \sum \infty [\frac{2}{5 ^{n}}]

n = 1 \sum \infty 3 (- \frac{1}{5})^{3 n}

n = 2 \sum \infty \frac{e ^{\frac{1}{n}}}{n}

What is the sum from n=0 to infinity of-ln(3-n) ?
The sum from n=0 to infinity of-ln(3-n) is diverges