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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{\partial}{\partial x} (sec^{2} (x y))

n = 1 \sum \infty \frac{3}{π ^{n}}

x^{\frac{1}{3}} - x^{- \frac{2}{3}}

x \to + 0 lim (2 x^{2} cos (\frac{5}{x}))

\frac{\partial}{\partial x} (e^{x} - 2 - cos (x y))

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