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

The sum from n=1 to infinity of 2sin(2/n) is diverges

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sum from n=1 to infinity of 2sin(2/n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply Series Limit Comparison Test:

diverges

= 2 diverges

= diverges

n = 1 \sum \infty \frac{1}{( n + 3 ) !}

n = 0 \sum \infty \frac{1}{n ln ^{2} ( n )}

n = 1 \sum \infty 70 (0.1)^{n}

n = 1 \sum \infty \frac{8 ^{n}}{7 ^{n}}

n = 6 \sum \infty ln (1 + \frac{1}{n})

What is the sum from n=1 to infinity of 2sin(2/n) ?
The sum from n=1 to infinity of 2sin(2/n) is diverges