What is the sum from n=1 to infinity of 6cos(1/(3n)) ?

The sum from n=1 to infinity of 6cos(1/(3n)) is diverges

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sum from n=1 to infinity of 6cos(1/(3n))

n = 1 \sum \infty 6 cos (\frac{1}{3 n})

diverges

Solution steps

n = 1 \sum \infty 6 cos (\frac{1}{3 n})

Apply the constant multiplication rule:

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

= 6 \cdot n = 1 \sum \infty cos (\frac{1}{3 n})

Apply Series Divergence Test:

diverges

= 6 diverges

= diverges

n = 0 \sum \infty \frac{n}{4 + n ^{2}}

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

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

n = 1 \sum \infty \frac{2}{n ^{2} + n ^{3}}

n = 1 \sum \infty \frac{9 ^{n}}{n ^{5}}

What is the sum from n=1 to infinity of 6cos(1/(3n)) ?
The sum from n=1 to infinity of 6cos(1/(3n)) is diverges