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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{n !}{( 3 n ) !}

n = 1 \sum \infty sec (\frac{1}{n})

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

n = 0 \sum \infty ln (n + 1)

n = 0 \sum \infty \frac{1}{9 + n}

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