What is the sum from n=1 to infinity of n^{-1/5} ?

The sum from n=1 to infinity of n^{-1/5} is diverges

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sum from n=1 to infinity of n^{-1/5}

n = 1 \sum \infty n^{- \frac{1}{5}}

diverges

Solution steps

n = 1 \sum \infty n^{- \frac{1}{5}}

Apply exponent rule:

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

= n = 1 \sum \infty \frac{1}{n ^{\frac{1}{5}}}

Apply p-Series Test:

diverges

= diverges

x \to 2 - lim (\frac{5 x}{x - 6})

f (x) = (x^{3} + 2) e^{x}

x \to - 2 lim (x^{2} - 4 x + 6)

ln (2 x - 5)

arctan (s)

What is the sum from n=1 to infinity of n^{-1/5} ?
The sum from n=1 to infinity of n^{-1/5} is diverges