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

The sum from n=0 to infinity of 2(1/2)^n is 4

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

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

4

Solution steps

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

Apply the constant multiplication rule:

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

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

Apply Series Geometric Test:

2

= 2 \cdot 2

Simplify

= 4

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

n = 1 \sum \infty - \frac{1}{1 1 ^{n}}

n = 1 \sum \infty 0.001 n

n = 1 \sum \infty \frac{10}{n ^{3}}

n = 0 \sum \infty 5^{- 2 n + 1} 3 n

What is the sum from n=0 to infinity of 2(1/2)^n ?
The sum from n=0 to infinity of 2(1/2)^n is 4