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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

diverges

= 2 diverges

= diverges

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}

n = 1 \sum \infty 3 (0.1)^{n - 1}

n = 1 \sum \infty 6 (0.4)^{n - 1}

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