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

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

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

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

diverges

Solution steps

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

Apply Series Limit Comparison Test:

diverges

= diverges

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d er i v a t i v e 6 (5)^{2} - 2 (5) + 9

\frac{d}{d x} (e^{2 x} + x^{2} - 16)

\int_{0}^{2} (1 + 3 x^{2}) d x

in v erse l a pl a ce \frac{s ^{2} + 8}{( s + 2 ) ^{4}}

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