What is the sum from n=1 to infinity of n^{(-2/3)} ?

The sum from n=1 to infinity of n^{(-2/3)} is diverges

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sum from n=1 to infinity of n^{(-2/3)}

n = 1 \sum \infty n^{(- \frac{2}{3})}

diverges

Solution steps

n = 1 \sum \infty n^{- \frac{2}{3}}

Apply exponent rule:

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

= n = 1 \sum \infty \frac{1}{n ^{\frac{2}{3}}}

Apply p-Series Test:

diverges

= diverges

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

x - 2

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

n = 1.0 E 22 \sum \infty \frac{1}{n}

\int_{0}^{1} (3 x + 1)^{2} d x

What is the sum from n=1 to infinity of n^{(-2/3)} ?
The sum from n=1 to infinity of n^{(-2/3)} is diverges