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

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

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

\int \frac{cos ( 6 x )}{sin ( 6 x )} d x

\int 2 x - 6 x^{3} d x

\frac{d}{d x} (2 π - x cos (x))

cos^{2} (x) sin (x) \frac{d y}{d x} cos^{3} (x) y = 1

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