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

x \to 0 + lim (\frac{x ^{4 x}}{7 x})

\int u^{- \frac{2}{3}} d u

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\frac{d}{d x} (\frac{x ^{3} + 1}{x ^{2} - x - 2})

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

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