What is the sum from n=0 to infinity of 5(-4/3)^n ?

The sum from n=0 to infinity of 5(-4/3)^n is diverges

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sum from n=0 to infinity of 5(-4/3)^n

n = 0 \sum \infty 5 (- \frac{4}{3})^{n}

diverges

Solution steps

n = 0 \sum \infty 5 (- \frac{4}{3})^{n}

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty (- \frac{4}{3})^{n}

Apply Series Geometric Test:

diverges

= 5 diverges

= diverges

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2 x^{- \frac{3}{4}}

\frac{d y}{d x} = (1 + e^{x^{2}}) (5 + x)

What is the sum from n=0 to infinity of 5(-4/3)^n ?
The sum from n=0 to infinity of 5(-4/3)^n is diverges