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

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

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

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

converges

Solution steps

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

Apply Series Ratio Test:

converges

= converges

n = 0 \sum \infty \frac{n}{( 1 + 4 n ) ^{2}}

n = 1 \sum \infty cos (8 n)

n = 0 \sum \infty n^{^{'}}

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

n = 1 \sum \infty 2 (\frac{3}{4})^{n - 1}

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