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

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

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

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

diverges

Solution steps

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

Apply Series Ratio Test:

diverges

= diverges

\frac{1}{x ^{2} + 1} (\frac{d y}{d x}) = x e^{y}

\int \frac{4}{x ^{2} - 1} d x

\frac{d}{d x} (- \frac{1}{2} sin (x))

\int \frac{x ^{2}}{a ^{2} - x ^{2}} d x

e^{x}, e^{- x}, [- 1, 1]

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