What is the sum from n=1 to infinity of 5cos(2/n) ?

The sum from n=1 to infinity of 5cos(2/n) is diverges

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sum from n=1 to infinity of 5cos(2/n)

n = 1 \sum \infty 5 cos (\frac{2}{n})

diverges

Solution steps

n = 1 \sum \infty 5 cos (\frac{2}{n})

Apply the constant multiplication rule:

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

= 5 \cdot n = 1 \sum \infty cos (\frac{2}{n})

Apply Series Divergence Test:

diverges

= 5 diverges

= diverges

6 x^{2} sin (x) tan (x)

\frac{d y}{d x}, x = y^{2}

\frac{d}{d x} (\frac{400}{x})

\frac{d y}{d x} = \frac{x ^{2} - 7}{15 y ^{2}}

\frac{d}{d x} (3 x^{2} + 4 x - 1)

What is the sum from n=1 to infinity of 5cos(2/n) ?
The sum from n=1 to infinity of 5cos(2/n) is diverges