What is the sum from n=8 to infinity of 1/n sin(1/n) ?

The sum from n=8 to infinity of 1/n sin(1/n) is converges

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sum from n=8 to infinity of 1/n sin(1/n)

n = 8 \sum \infty \frac{1}{n} sin (\frac{1}{n})

converges

Solution steps

n = 8 \sum \infty \frac{1}{n} sin (\frac{1}{n})

Apply Series Integral Test:

converges

= converges

y^{^{'}} = \frac{t y}{t + 1}, y (4) = 1

(x + y) d x + (3 x + 3 y - 4) d y = 0

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

f (x) = 6 (ln (x + 1))

What is the sum from n=8 to infinity of 1/n sin(1/n) ?
The sum from n=8 to infinity of 1/n sin(1/n) is converges