What is the sum from n=1 to infinity of n^{-0.3} ?

The sum from n=1 to infinity of n^{-0.3} is diverges

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sum from n=1 to infinity of n^{-0.3}

n = 1 \sum \infty n^{- 0.3}

diverges

Solution steps

n = 1 \sum \infty n^{- 0.3}

Apply Series Integral Test:

diverges

= diverges

n = 1 \sum \infty \frac{sin ( nπ )}{n}

x = 0 \sum \infty \frac{x}{2 ^{x}}

x = 0 \sum \infty \frac{1}{x !}

n = 4 \sum \infty \frac{2}{n ( n - 1 )}

n = 1 \sum \infty \frac{n}{n ^{2} + 7}

What is the sum from n=1 to infinity of n^{-0.3} ?
The sum from n=1 to infinity of n^{-0.3} is diverges