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

2 x + \frac{3}{x}

\frac{\partial}{\partial x} (2 x + 5 y + 3)

(y + 1) \frac{d y}{d x} = 2 x

x^{3}, x^{5}

\int_{2}^{5} (1 - x^{2}) - (11 - 7 x) d x

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