What is the sum from n=0 to infinity of sin(3n) ?

The sum from n=0 to infinity of sin(3n) is diverges

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sum from n=0 to infinity of sin(3n)

n = 0 \sum \infty sin (3 n)

diverges

Solution steps

n = 0 \sum \infty sin (3 n)

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{1}{( 1 + n ) ^{4}}

n = 1 \sum \infty (\frac{4}{9})^{n - 1}

n = 0 \sum \infty (3.5)^{- n}

n = 1 \sum \infty \frac{10 n}{4 n + 1}

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

What is the sum from n=0 to infinity of sin(3n) ?
The sum from n=0 to infinity of sin(3n) is diverges