What is the sum from n=1 to infinity of n^{-1/9} ?

The sum from n=1 to infinity of n^{-1/9} is diverges

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sum from n=1 to infinity of n^{-1/9}

n = 1 \sum \infty n^{- \frac{1}{9}}

diverges

Solution steps

n = 1 \sum \infty n^{- \frac{1}{9}}

Apply exponent rule:

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

= n = 1 \sum \infty \frac{1}{n ^{\frac{1}{9}}}

Apply p-Series Test:

diverges

= diverges

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

x \to 9 lim (\frac{( x + 2 )}{2})

x y d x + 3 e^{x} d y = 0

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

\frac{1}{( 1 + 4 x )}

What is the sum from n=1 to infinity of n^{-1/9} ?
The sum from n=1 to infinity of n^{-1/9} is diverges