What is the sum from n=0 to infinity of 3/(2n) ?

The sum from n=0 to infinity of 3/(2n) is diverges

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sum from n=0 to infinity of 3/(2n)

n = 0 \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

n = 0 \sum \infty \frac{1}{6} (\frac{5}{6})^{n}

n = 1 \sum \infty - n^{- \frac{1}{3}}

n = 1 \sum \infty \frac{10 0 ^{n}}{n !}

n = 1 \sum \infty n ln (n)

k = 2 \sum \infty \frac{1}{k ^{2} - 1}

What is the sum from n=0 to infinity of 3/(2n) ?
The sum from n=0 to infinity of 3/(2n) is diverges