What is the sum from n=1 to infinity of 4(0.5)^{n-1} ?

The sum from n=1 to infinity of 4(0.5)^{n-1} is converges

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sum from n=1 to infinity of 4(0.5)^{n-1}

n = 1 \sum \infty 4 (0.5)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 4 \cdot 0. 5^{n - 1}

Apply the constant multiplication rule:

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

= 4 \cdot n = 1 \sum \infty 0. 5^{n - 1}

Apply Series Ratio Test:

converges

= 4 converges

= converges

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

n = 1 \sum \infty \frac{1}{16 n ^{2} - 8 n - 3}

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

n = 1 \sum \infty \frac{5}{( - 4 ) ^{n}}

n = 1 \sum \infty \frac{n !}{11 4 ^{n}}

What is the sum from n=1 to infinity of 4(0.5)^{n-1} ?
The sum from n=1 to infinity of 4(0.5)^{n-1} is converges