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

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

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

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

converges

Solution steps

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

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 - 1}}

Apply Series Ratio Test:

converges

= 3 converges

= converges

n = 1 \sum \infty \frac{5 ^{n}}{8}

n = 0 \sum \infty \frac{n}{49 + n ^{2}}

k = 2 \sum \infty ln (\frac{k}{k + 1})

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

n = 5 \sum \infty e^{5 - 6 n}

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