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

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

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

n = 1 \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

\int (4 t^{2} - 3)^{2} d t

f (x) = x^{2}, [1, 5]

\frac{8 x ^{2} + 2}{x ^{2}}

(x^{2} - 4 x + 5) 25 - x^{2}

\int t - 2 cos (t) d t

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