What is the sum from n=0 to infinity of sin(2n) ?

The sum from n=0 to infinity of sin(2n) is diverges

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sum from n=0 to infinity of sin(2n)

n = 0 \sum \infty sin (2 n)

diverges

Solution steps

n = 0 \sum \infty sin (2 n)

Apply Series Divergence Test:

diverges

= diverges

\frac{x + x}{x ^{2}}

\frac{\partial}{\partial x} (4 x ln (x y))

\int x^{11} d x

y^{^{''}} + 4 y = 1

y^{^{'}} + y = - cos (6 x)

What is the sum from n=0 to infinity of sin(2n) ?
The sum from n=0 to infinity of sin(2n) is diverges