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

f (x) = cot^{3} (4 x)

(\frac{( x + 2 )}{x - 2})^{2}

e^{5 x}, e^{10 x}, x = 1

(x + 2) (10 - 3 x)

\int \frac{1}{x ^{2} - 6 x + 12} d x

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