What is the sum from n=1 to infinity of 1/(n^{3/2)} ?

The sum from n=1 to infinity of 1/(n^{3/2)} is converges

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sum from n=1 to infinity of 1/(n^{3/2)}

n = 1 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

converges

Solution steps

n = 1 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

Apply p-Series Test:

converges

= converges

\frac{d}{d x} (\frac{87 x}{99 x - 87})

\frac{s + 1}{s ^{2} - 5 s + 6}

y = x^{3} - 6 x^{2} + 9 x, y = x

\frac{\partial}{\partial x} (- x (x^{2} + y^{2} + z^{2})^{- \frac{3}{2}})

\int 100 - x d x

What is the sum from n=1 to infinity of 1/(n^{3/2)} ?
The sum from n=1 to infinity of 1/(n^{3/2)} is converges