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

The sum from n=2 to infinity of ln(1-1/(n^2)) is converges (absolutely)

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

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

converges (absolutely)

Solution steps

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

The series has positive and negative elements, check for conditional/absolute convergence

Check convergence of

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

converges

= converges (absolutely)

y = \frac{x ^{2}}{x - 8}, (2, \frac{2}{5})

\int sin (e^{2 x}) d x

\frac{5 x}{x - 2}

f (x) = 6 x^{2} - 3, at x = - 9

\frac{( s ^{2} )}{( s + 2 ) ^{4}}

What is the sum from n=2 to infinity of ln(1-1/(n^2)) ?
The sum from n=2 to infinity of ln(1-1/(n^2)) is converges (absolutely)