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

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

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

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

diverges

Solution steps

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

Apply Series Geometric Test:

diverges

= diverges

x + \frac{1}{x ^{3}}

\int \frac{1}{64 e ^{- 6 x} + e ^{6 x}} d x

\frac{d}{d x} (2 sec (\frac{x}{2}))

\int \frac{1}{3 + x} d x

2 y^{^{'}} + 5 y = 4

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