What is the sum from n=4 to infinity of 3(-1/2)^n ?

The sum from n=4 to infinity of 3(-1/2)^n is converges

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sum from n=4 to infinity of 3(-1/2)^n

n = 4 \sum \infty 3 (- \frac{1}{2})^{n}

converges

Solution steps

n = 4 \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 = 4 \sum \infty (- \frac{1}{2})^{n}

Apply Series Ratio Test:

converges

= 3 converges

= converges

\frac{a + x}{a ^{2} + x ^{2}}

2 x - 1

x \to 0 lim (\frac{1}{∣ x ∣})

- 2 x^{2} + 8, 3 x + 1, x = 2, x = 0

f (x) = 2 x e^{x} + e^{x} x^{2}

What is the sum from n=4 to infinity of 3(-1/2)^n ?
The sum from n=4 to infinity of 3(-1/2)^n is converges