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

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

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

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

diverges

Solution steps

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

Simplify

= n = 0 \sum \infty sin (1)

Every infinite sum of a non-zero constant diverges

= diverges

n = 1 \sum \infty \frac{5}{n + 4}

n = 2 \sum \infty (- 0.95)^{n}

n = 1 \sum \infty 10 (- \frac{3}{8})^{n}

n = 1 \sum \infty \frac{6 ^{n}}{n ^{6}}

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

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