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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

\int_{1}^{2} (x + \frac{1}{x})^{2} d x

\frac{\partial}{\partial y} (5 x^{2} - 3 x y + x yz)

x \to \infty lim (- 2 \cdot x)

\frac{d}{d x} (3 x^{5} + 8 x^{2})

y = e^{- x}, x = 2, y = 1

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