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

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

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

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

converges

Solution steps

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

Simplify

\frac{n}{n ^{2}} : \frac{1}{n ^{\frac{3}{2}}}

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

Apply p-Series Test:

converges

= converges

\int_{0}^{1} ln (3 x) d x

\int \frac{2 x}{( 2 + 3 x ^{2} ) ^{3}} d x

\int_{0}^{4} \frac{1}{x ^{2} + 16} d x

\frac{x}{4 + x}

\frac{\partial}{\partial x} (x y + x e^{\frac{y}{x}})

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