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

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

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

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

converges

Solution steps

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

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

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

Apply p-Series Test:

converges

= 2 converges

= converges

x y^{2} y^{^{'}} = x + 1

\int \frac{9 - 4 x}{x} d x

y = 9 e^{x} + x, (0, 9)

\frac{s + 5}{s ( s + 2 ) ( s + 3 )}

\int e^{0}

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