What is the sum from n=0 to infinity of 1/(sin(n)) ?

The sum from n=0 to infinity of 1/(sin(n)) is diverges

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sum from n=0 to infinity of 1/(sin(n))

n = 0 \sum \infty \frac{1}{sin ( n )}

diverges

Solution steps

n = 0 \sum \infty \frac{1}{sin ( n )}

Apply Series Divergence Test:

diverges

= diverges

x \to \frac{π}{6} lim (x sin (x))

y^{^{'}} = 3 y

y = x^{2} + 2, (- 3, 11)

x \to 1 lim (\frac{( 26 - x - 5 )}{17 - x - 4})

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{3 t}

What is the sum from n=0 to infinity of 1/(sin(n)) ?
The sum from n=0 to infinity of 1/(sin(n)) is diverges