What is the sum from n=1 to infinity of 1/(\sqrt[n]{2)} ?

The sum from n=1 to infinity of 1/(\sqrt[n]{2)} is diverges

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sum from n=1 to infinity of 1/(\sqrt[n]{2)}

diverges

Solution steps

Apply radical rule:

a = a^{\frac{1}{2}}

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply Series Divergence Test:

diverges

= diverges

y^{^{''}} + 9 y = (sec (3 t))^{2}

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

t e^{- t}

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

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

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