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

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

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

n = 1 \sum \infty n^{- s i n (1)}

diverges

Solution steps

n = 1 \sum \infty n^{- s i n (1)}

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty cos^{n} (5)

k = 1 \sum \infty (\frac{2}{5})^{k}

n = 5 \sum \infty \frac{1}{ln ( n - 3 )}

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

n = 0 \sum \infty \frac{5}{8 ^{n}}

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