What is the sum from k=1 to infinity of 1/(2+sin(k)) ?

The sum from k=1 to infinity of 1/(2+sin(k)) is diverges

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sum from k=1 to infinity of 1/(2+sin(k))

k = 1 \sum \infty \frac{1}{2 + sin ( k )}

diverges

Solution steps

k = 1 \sum \infty \frac{1}{2 + sin ( k )}

Apply Series Divergence Test:

diverges

= diverges

n = 1 \sum \infty \frac{6}{n ^{2} + 6 n + 8}

n = 0 \sum \infty \frac{n}{5 n ^{3} + 1}

n = 0 \sum \infty 9^{\frac{1}{n}}

n = 0 \sum \infty 3 (\frac{2}{5})^{n + 1}

n = 0 \sum \infty \frac{4}{3 + n}

What is the sum from k=1 to infinity of 1/(2+sin(k)) ?
The sum from k=1 to infinity of 1/(2+sin(k)) is diverges