What is the sum from n=1 to infinity of 12sin(1/n) ?

The sum from n=1 to infinity of 12sin(1/n) is diverges

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sum from n=1 to infinity of 12sin(1/n)

n = 1 \sum \infty 12 sin (\frac{1}{n})

diverges

Solution steps

n = 1 \sum \infty 12 sin (\frac{1}{n})

Apply the constant multiplication rule:

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

= 12 \cdot n = 1 \sum \infty sin (\frac{1}{n})

Apply Series Limit Comparison Test:

diverges

= 12 diverges

= diverges

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

n = 0 \sum \infty n^{- \frac{3}{2}}

n = 3 \sum \infty \frac{1}{( n - 2 ) ^{4}}

n = 1 \sum \infty \frac{9}{7 ^{n}}

n = 5 \sum \infty \frac{4}{n ^{2} - n}

What is the sum from n=1 to infinity of 12sin(1/n) ?
The sum from n=1 to infinity of 12sin(1/n) is diverges