What is the sum from n=0 to infinity of (3^n)/(1000) ?

The sum from n=0 to infinity of (3^n)/(1000) is diverges

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sum from n=0 to infinity of (3^n)/(1000)

n = 0 \sum \infty \frac{3 ^{n}}{1000}

diverges

Solution steps

n = 0 \sum \infty \frac{3 ^{n}}{1000}

Apply the constant multiplication rule:

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

= \frac{1}{1000} \cdot n = 0 \sum \infty 3^{n}

Apply Series Geometric Test:

diverges

= \frac{1}{1000} diverges

= diverges

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\frac{d}{d x} (\frac{4 x - 4 1 + x}{x ^{2} - 1})

y = 6 cos (x) + 3 sin (2 x)

\frac{d}{d x} (\frac{8 x ^{2} + 8 x + 2}{x})

\int \frac{4}{x ^{2} 16 - x ^{2}} d x

What is the sum from n=0 to infinity of (3^n)/(1000) ?
The sum from n=0 to infinity of (3^n)/(1000) is diverges