What is the sum from n=1 to infinity of ((4pi)/5)^n ?

The sum from n=1 to infinity of ((4pi)/5)^n is diverges

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sum from n=1 to infinity of ((4pi)/5)^n

n = 1 \sum \infty (\frac{4 π}{5})^{n}

diverges

Solution steps

n = 1 \sum \infty (\frac{4 π}{5})^{n}

Apply Series Ratio Test:

diverges

= diverges

\int x^{2} + x - 3 d x

\frac{d}{d x} (sin (4 x) + sin^{2} (4 x))

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} x}{d t} + 2 \frac{d x}{d t} + x = 1

\int 5 + x^{2} d x

What is the sum from n=1 to infinity of ((4pi)/5)^n ?
The sum from n=1 to infinity of ((4pi)/5)^n is diverges