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}

(- \frac{3}{2})^{n} = 1. 5^{n} (- 1)^{n}

= 5 \cdot n = 0 \sum \infty 1. 5^{n} (- 1)^{n}

Apply Series Ratio Test:

diverges

= 5 diverges

= diverges

n = 15 \sum \infty (\frac{2}{3})^{n}

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

n = 1 \sum \infty 12 sin^{n} (7 0^{\circ})

n = 0 \sum \infty \frac{n}{5 n ^{2} + 1}

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

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