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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\frac{d}{d x} ((x^{3} - 2)^{2})

\int 1000 t e^{- 0.5 t} d t

y^{^{''}} - 4 y^{^{'}} + 4 y = 120 x^{2} e^{2 x}

x \to - 4 lim (\frac{x ^{2} + 5}{x + 4})

3 x^{2} y^{^{'}} = y^{^{'}} + 3 x e^{- y}

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