What is the sum from n=1 to infinity of 1/(n^{4/5)} ?

The sum from n=1 to infinity of 1/(n^{4/5)} is diverges

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sum from n=1 to infinity of 1/(n^{4/5)}

n = 1 \sum \infty \frac{1}{n ^{\frac{4}{5}}}

diverges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{\frac{4}{5}}}

Apply exponent rule:

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

\frac{1}{n ^{\frac{4}{5}}} = n^{- \frac{4}{5}}

= n = 1 \sum \infty n^{- \frac{4}{5}}

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty \frac{1}{9 n ^{2} + 23}

m = 2 \sum \infty \frac{6}{7 ^{m}}

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

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

n = 0 \sum \infty (\frac{- 7}{8})^{n}

What is the sum from n=1 to infinity of 1/(n^{4/5)} ?
The sum from n=1 to infinity of 1/(n^{4/5)} is diverges