What is the sum from n=1 to infinity of 3/(2n) ?

The sum from n=1 to infinity of 3/(2n) is diverges

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sum from n=1 to infinity of 3/(2n)

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

= \frac{3}{2} \cdot n = 1 \sum \infty \frac{1}{n}

Apply Cauchy's Convergence Condition:

diverges

= \frac{3}{2} diverges

= diverges

k = 1 \sum \infty k e^{- 7 k}

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

k = 0 \sum \infty (\frac{2}{3})^{k}

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

k = 1 \sum \infty \frac{1}{k ^{2} + 5 k + 6}

What is the sum from n=1 to infinity of 3/(2n) ?
The sum from n=1 to infinity of 3/(2n) is diverges