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

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

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

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

diverges

Solution steps

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

Apply Series Divergence Test:

diverges

= diverges

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

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

\int x e^{- 6 x^{2}} d x

x \to - 2 lim (4 x^{2} - 3 x + 5)

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

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