What is the sum from n=1 to infinity of n^{-0.8} ?

The sum from n=1 to infinity of n^{-0.8} is diverges

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sum from n=1 to infinity of n^{-0.8}

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

diverges

Solution steps

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

Apply exponent rule:

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

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

Apply p-Series Test:

diverges

= diverges

n = 2 \sum \infty \frac{4}{n ( n + 2 )}

k = 1 \sum \infty \frac{k}{e ^{k}}

n = 0 \sum \infty (1 + \frac{3}{n})^{n}

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

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

What is the sum from n=1 to infinity of n^{-0.8} ?
The sum from n=1 to infinity of n^{-0.8} is diverges