What is the sum from n=1 to infinity of n^{-sin(2)} ?

The sum from n=1 to infinity of n^{-sin(2)} is diverges

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sum from n=1 to infinity of n^{-sin(2)}

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

diverges

Solution steps

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

Apply Series Integral Test:

diverges

= diverges

(cos^{2} (θ))^{^{'}}

\frac{1}{1 - x}

\frac{d}{d x} (\frac{8 x}{3 - tan ( x )})

f (x) = 7 x + 2

\frac{d y}{d x} = \frac{( x - 3 )}{2 - y}

What is the sum from n=1 to infinity of n^{-sin(2)} ?
The sum from n=1 to infinity of n^{-sin(2)} is diverges