What is the sum from n=0 to infinity of n^{-1/2} ?

The sum from n=0 to infinity of n^{-1/2} is diverges

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sum from n=0 to infinity of n^{-1/2}

n = 0 \sum \infty n^{- \frac{1}{2}}

diverges

Solution steps

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

Apply exponent rule:

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

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

Apply p-Series Test:

diverges

= diverges

x = 1 \sum \infty \frac{ln ( x )}{x}

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

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

n = 0 \sum \infty 3 + \frac{2}{n ^{4}}

n = 1 \sum \infty \frac{3}{n ^{9}}

What is the sum from n=0 to infinity of n^{-1/2} ?
The sum from n=0 to infinity of n^{-1/2} is diverges