What is the sum from n=0 to infinity of 3(-1/2)^n ?

The sum from n=0 to infinity of 3(-1/2)^n is converges

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sum from n=0 to infinity of 3(-1/2)^n

n = 0 \sum \infty 3 (- \frac{1}{2})^{n}

converges

Solution steps

n = 0 \sum \infty 3 (- \frac{1}{2})^{n}

Apply the constant multiplication rule:

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

= 3 \cdot n = 0 \sum \infty (- \frac{1}{2})^{n}

Apply Series Ratio Test:

converges

= 3 converges

= converges

x \to 2 lim (\frac{4}{( x - 2 ) ^{5}})

f (x) = x + e^{x} - \frac{1}{x}, at x = 1

\int - \frac{2 x}{x ^{4} - 1} d x

x \to \frac{π}{3} lim (cos (x))

x \to 1 lim (\frac{- x + 1}{x - 1})

What is the sum from n=0 to infinity of 3(-1/2)^n ?
The sum from n=0 to infinity of 3(-1/2)^n is converges