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

2 (t - 3)^{2} + 9 (t - 3) + 11

y = 3 e^{4 x} + 6 e^{2 x}

\frac{d}{d x} (e^{7 x} + e^{- 7 x})

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

f (x) = 20 x - 16 x^{2}, at x = 3

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