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

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

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sum from n=0 to infinity of n^{-sin(1)}

n = 0 \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 \frac{2 n}{7 n - 3}

n = 0 \sum \infty \frac{1}{n ^{0}}

n = 2 \sum \infty \frac{7}{n ( n + 1 )}

n = 0 \sum \infty 0.2 5^{n}

n = 1 \sum \infty sin^{n} (3)

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