What is the sum from n=0 to infinity of ln(2-n) ?

The sum from n=0 to infinity of ln(2-n) is diverges

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sum from n=0 to infinity of ln(2-n)

n = 0 \sum \infty ln (2 - n)

diverges

Solution steps

n = 3 \sum \infty ln (2 - n)

Apply Series Divergence Test:

diverges

= diverges

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

n = 4 \sum \infty \frac{1}{n ^{2} - 4 n + 3}

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

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

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

What is the sum from n=0 to infinity of ln(2-n) ?
The sum from n=0 to infinity of ln(2-n) is diverges