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

k = 2 \sum \infty (\frac{3}{8})^{2 k}

n = 2 \sum \infty arctan (n)

n = 0 \sum \infty ln (1 + 5 n)

n = 0 \sum \infty 3 (\frac{1}{11})^{n}

n = 1 \sum \infty (- \frac{6}{8})^{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