What is the sum from n=1 to infinity}((\sqrt{n of+4))/(n^2) ?

The sum from n=1 to infinity}((\sqrt{n of+4))/(n^2) is converges

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sum from n=1 to infinity}((\sqrt{n of+4))/(n^2)

n = 1 \sum \infty \frac{( n + 4 )}{n ^{2}}

converges

Solution steps

n = 1 \sum \infty \frac{n + 4}{n ^{2}}

Expand

\frac{n + 4}{n ^{2}} : \frac{1}{n ^{\frac{3}{2}}} + \frac{4}{n ^{2}}

= n = 1 \sum \infty \frac{1}{n ^{\frac{3}{2}}} + \frac{4}{n ^{2}}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 1 \sum \infty \frac{1}{n ^{\frac{3}{2}}} + n = 1 \sum \infty \frac{4}{n ^{2}}

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

converges

n = 1 \sum \infty \frac{4}{n ^{2}} =

converges

= converges + converges

= converges

x \to 3 lim (2 x^{2} - 2)

\frac{d y}{d x} + y^{7} x + y = 0

\frac{d}{d x} (\frac{3 cos ( x )}{1 - 2 sin ( x )})

\frac{\partial}{\partial v} (\frac{u - v}{3})

\int \frac{e ^{x}}{e ^{x} + 1} d x

What is the sum from n=1 to infinity}((\sqrt{n of+4))/(n^2) ?
The sum from n=1 to infinity}((\sqrt{n of+4))/(n^2) is converges