What is the sum from k=1 to infinity of 1/(2^{k+1)} ?

The sum from k=1 to infinity of 1/(2^{k+1)} is converges

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sum from k=1 to infinity of 1/(2^{k+1)}

k = 1 \sum \infty \frac{1}{2 ^{k + 1}}

converges

Solution steps

k = 1 \sum \infty \frac{1}{2 ^{k + 1}}

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{4}{9 ^{n}}

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

n = 1 \sum \infty n (\frac{8}{9})^{n}

k = 1 \sum \infty \frac{1}{k} - \frac{1}{k + 2}

n = 1 \sum \infty 6 ln (\frac{n}{n + 1})

What is the sum from k=1 to infinity of 1/(2^{k+1)} ?
The sum from k=1 to infinity of 1/(2^{k+1)} is converges