What is the sum from n=1 to infinity of 4/(2n) ?

The sum from n=1 to infinity of 4/(2n) is diverges

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sum from n=1 to infinity of 4/(2n)

n = 1 \sum \infty \frac{4}{2 n}

diverges

Solution steps

n = 1 \sum \infty \frac{4}{2 n}

Apply the constant multiplication rule:

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

= \frac{4}{2} \cdot n = 1 \sum \infty \frac{1}{n}

Simplify

= 2 \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= 2 diverges

= diverges

\frac{d ^{3}}{d x ^{3}} (\frac{6}{x ^{2}})

\int (y^{2} cos (x) - 3 x^{2} y - 2 x) d x

\frac{\partial}{\partial z} (ln (x y))

x^{3} - 4 x^{2}, 2 x^{2} - 8 x

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 3 s )}

What is the sum from n=1 to infinity of 4/(2n) ?
The sum from n=1 to infinity of 4/(2n) is diverges