What is the sum from n=0 to infinity of 5/(2^{n-1)} ?

The sum from n=0 to infinity of 5/(2^{n-1)} is converges

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sum from n=0 to infinity of 5/(2^{n-1)}

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

converges

Solution steps

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

Apply the constant multiplication rule:

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

= 5 \cdot n = 0 \sum \infty \frac{1}{2 ^{n - 1}}

Apply Series Ratio Test:

converges

= 5 converges

= converges

n = 0 \sum \infty \frac{n}{1 - 4 n}

n = 16 \sum \infty \frac{1}{ln ( n )}

n = 1 \sum \infty \frac{1}{9 n}

k = 2 \sum \infty (\frac{3}{5})^{3 k}

n = 4 \sum \infty \frac{1}{n}

What is the sum from n=0 to infinity of 5/(2^{n-1)} ?
The sum from n=0 to infinity of 5/(2^{n-1)} is converges