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

The sum from n=1 to infinity of 1-1/(2^n) is diverges

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sum from n=1 to infinity of 1-1/(2^n)

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

n = 0 \sum \infty 5^{n} 4^{- n}

k = 1 \sum \infty 5^{- k}

n = 1 \sum \infty n (\frac{- 5}{9})^{n}

n = 1 \sum \infty \frac{2}{n ^{2} + 6 n}

What is the sum from n=1 to infinity of 1-1/(2^n) ?
The sum from n=1 to infinity of 1-1/(2^n) is diverges