What is the sum from n=5 to infinity of sin^2(n) ?

The sum from n=5 to infinity of sin^2(n) is diverges

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sum from n=5 to infinity of sin^2(n)

n = 5 \sum \infty sin^{2} (n)

diverges

Solution steps

n = 5 \sum \infty sin^{2} (n)

Apply Series Divergence Test:

diverges

= diverges

\int 4 sin (3 t) + 1.2 e^{- 6 t} d t

y = \frac{2}{x}, (- 4, - \frac{1}{2})

\int \frac{3 x}{16 - 9 x ^{2}} d x

x \to 0 + lim (x ln (x^{4}))

What is the sum from n=5 to infinity of sin^2(n) ?
The sum from n=5 to infinity of sin^2(n) is diverges