What is the sum from n=1 to infinity of 6sin(1/n) ?

The sum from n=1 to infinity of 6sin(1/n) is diverges

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sum from n=1 to infinity of 6sin(1/n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty sin (\frac{1}{n})

Apply Series Limit Comparison Test:

diverges

= 6 diverges

= diverges

\frac{12 ( s + 1 )}{s ( s ^{2} + 9 )}

\frac{\partial}{\partial x} (cos (3 y))

\frac{d}{d x} (\frac{x}{x - 6})

\int \frac{1}{4} x^{2} d x

\int 2 x (x^{2} + 4)^{2} d x

What is the sum from n=1 to infinity of 6sin(1/n) ?
The sum from n=1 to infinity of 6sin(1/n) is diverges