What is the sum from n=1 to infinity of 4(3/2)^{n-1} ?

The sum from n=1 to infinity of 4(3/2)^{n-1} is diverges

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sum from n=1 to infinity of 4(3/2)^{n-1}

n = 1 \sum \infty 4 (\frac{3}{2})^{n - 1}

diverges

Solution steps

n = 1 \sum \infty 4 (\frac{3}{2})^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty (\frac{3}{2})^{n - 1}

Apply Series Ratio Test:

diverges

= 4 diverges

= diverges

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\frac{d}{d x} (sin (x) \cdot cos (x) \cdot \frac{1}{x})

y = 6 x^{2}, y = 4 x^{2}, y = 6 - 5 x

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What is the sum from n=1 to infinity of 4(3/2)^{n-1} ?
The sum from n=1 to infinity of 4(3/2)^{n-1} is diverges