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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

n = 3 \sum \infty \frac{2}{3 ln ( n )}

n = 1 \sum \infty \frac{1}{9 n ^{4} + 3}

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

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

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