What is the sum from n=0 to infinity of sin(1/(n^1)) ?

The sum from n=0 to infinity of sin(1/(n^1)) is diverges

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sum from n=0 to infinity of sin(1/(n^1))

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

diverges

Solution steps

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

Simplify

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

Apply Series Limit Comparison Test:

diverges

= diverges

n = 0 \sum \infty (\frac{11}{12})^{n}

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

n = 1 \sum \infty \frac{2}{n ^{4}}

n = 1 \sum \infty \frac{2}{n ^{5}}

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

What is the sum from n=0 to infinity of sin(1/(n^1)) ?
The sum from n=0 to infinity of sin(1/(n^1)) is diverges