What is the sum from n=1 to infinity of sin(3n) ?

The sum from n=1 to infinity of sin(3n) is diverges

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sum from n=1 to infinity of sin(3n)

n = 1 \sum \infty sin (3 n)

diverges

Solution steps

n = 1 \sum \infty sin (3 n)

Apply Series Divergence Test:

diverges

= diverges

n = 0 \sum \infty \frac{2}{( - 3 ) ^{n}}

n = 1 \sum \infty \frac{2}{3 n - 1}

n = 2 \sum \infty \frac{- 1}{ln ( n )}

n = 3 \sum \infty \frac{7}{n} - \frac{7}{n + 2}

k = 0 \sum \infty \frac{k}{4 k + 1}

What is the sum from n=1 to infinity of sin(3n) ?
The sum from n=1 to infinity of sin(3n) is diverges