What is the sum from n=0 to infinity of 5(3/2)^n ?

The sum from n=0 to infinity of 5(3/2)^n is diverges

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sum from n=0 to infinity of 5(3/2)^n

n = 0 \sum \infty 5 (\frac{3}{2})^{n}

diverges

Solution steps

n = 0 \sum \infty 5 (\frac{3}{2})^{n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty (\frac{3}{2})^{n}

Apply Series Geometric Test:

diverges

= 5 diverges

= diverges

\int x^{2} 8 x + 9 d x

\frac{\partial}{\partial y} (\frac{y ^{3}}{x ^{2} + y ^{2}})

n = 0 \sum \infty \frac{3 n ^{9}}{n !}

\frac{d}{d x} (3 cos (x) + 3 sin (x))

\frac{d}{d x} (\frac{x}{x ^{2} + a ^{2}})

What is the sum from n=0 to infinity of 5(3/2)^n ?
The sum from n=0 to infinity of 5(3/2)^n is diverges