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

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

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

n = 1 \sum \infty sin (\frac{1}{n})

diverges

Solution steps

n = 1 \sum \infty sin (\frac{1}{n})

Apply Series Limit Comparison Test:

diverges

= diverges

\frac{\partial}{\partial x} (y^{2} sin (x z))

x \to 4 + lim (\frac{1}{x ^{3} - 64})

\int sec^{6} (2 x) d x

\int \frac{4}{5 + 2 x - x ^{2}} d x

3 (s^{3} - s^{2})

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