What is the sum from n=1 to infinity of (-1)^nsin(2/n) ?

The sum from n=1 to infinity of (-1)^nsin(2/n) is converges

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sum from n=1 to infinity of (-1)^nsin(2/n)

n = 1 \sum \infty (- 1)^{n} sin (\frac{2}{n})

converges

Solution steps

n = 1 \sum \infty (- 1)^{n} sin (\frac{2}{n})

Apply Alternating Series Test:

converges

= converges

\frac{d}{d x} (6 x + 4)

\int_{1}^{3} 3 r^{4} ln (r) d r

d y - (y - 5)^{2} d x = 0

x^{2}, 2 x^{4}, - 1, 1

x \to 0 lim (\frac{1}{sin ^{2} ( x )})

What is the sum from n=1 to infinity of (-1)^nsin(2/n) ?
The sum from n=1 to infinity of (-1)^nsin(2/n) is converges