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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

x \to 3 lim (\frac{x ^{2} - 1}{x - 3})

\frac{d v}{d t}

y^{2} \frac{d y}{d x} = y^{3} - x^{3}

x \to \frac{π}{2} lim (f (x))

\frac{d}{d x} (tan^{3} (4 x))

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