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

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

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

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

diverges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply exponent rule:

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

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

Apply p-Series Test:

diverges

= - diverges

= diverges

n = 1 \sum \infty \frac{10 0 ^{n}}{n !}

n = 1 \sum \infty n ln (n)

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

n = 0 \sum \infty \frac{cos ( nπ )}{n}

n = 1 \sum \infty n^{- 0.8}

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