What is the sum from n=1 to infinity of n^{-0.9999} ?

The sum from n=1 to infinity of n^{-0.9999} is diverges

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sum from n=1 to infinity of n^{-0.9999}

n = 1 \sum \infty n^{- 0.9999}

diverges

Solution steps

n = 1 \sum \infty n^{- 0.9999}

Apply exponent rule:

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

= n = 1 \sum \infty \frac{1}{n ^{0.9999}}

Apply p-Series Test:

diverges

= diverges

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

k = 0 \sum \infty \frac{4 ^{k}}{k !}

j = 1 \sum \infty 3 (\frac{1}{4})^{j}

n = 1 \sum \infty \frac{n}{n + 8}

n = 9 \sum \infty \frac{1}{9 n ^{4} + 8}

What is the sum from n=1 to infinity of n^{-0.9999} ?
The sum from n=1 to infinity of n^{-0.9999} is diverges