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

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

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

n = 1 \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

y = \frac{x ^{4}}{1 - x ^{3}}

x \to 5 + lim (\frac{x - 6}{x - 5})

\int 5 x^{- 6} d x

\frac{1}{8}

x \to 5 lim (\frac{9 x}{x ^{2} - 25})

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