What is the sum from n=0 to infinity of ln(1-n^2) ?

The sum from n=0 to infinity of ln(1-n^2) is diverges

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sum from n=0 to infinity of ln(1-n^2)

n = 0 \sum \infty ln (1 - n^{2})

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty 5^{- 2 n + 1} 3^{n}

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

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

k = 1 \sum \infty \frac{k}{2 ^{k - 1}}

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

What is the sum from n=0 to infinity of ln(1-n^2) ?
The sum from n=0 to infinity of ln(1-n^2) is diverges