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) sin (\frac{1}{n})

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

y^{^{'}} + a y - b = 0

\int \frac{5}{4 + 9 r ^{2}} d r

x \to \infty lim (4 + x)

\int \frac{( x + 3 )}{( x + 4 )} d x

\frac{d}{d x} (4 x^{3} + 3 x^{2} + 2 x + 1)

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