What is the sum from n=5 to infinity of (sin(n))/(n^{3/2)} ?

The sum from n=5 to infinity of (sin(n))/(n^{3/2)} is converges (absolutely)

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sum from n=5 to infinity of (sin(n))/(n^{3/2)}

n = 5 \sum \infty \frac{sin ( n )}{n ^{\frac{3}{2}}}

converges (absolutely)

Solution steps

n = 5 \sum \infty \frac{sin ( n )}{n ^{\frac{3}{2}}}

The series has positive and negative elements, check for conditional/absolute convergence

Check convergence of

n = 5 \sum \infty \frac{sin ( n )}{n ^{\frac{3}{2}}} :

converges

= converges (absolutely)

\int \frac{5}{x ^{5}} d x

\int 2 e^{c o s (x)} sin (x) d x

\int (3 x - \frac{1}{x} - 7 e^{x} + 14 e) d x

\int \frac{5 sec ^{2} ( x )}{2 ( 1 - tan ( x ) )} d x

\frac{d}{d x} (x y + 1)

What is the sum from n=5 to infinity of (sin(n))/(n^{3/2)} ?
The sum from n=5 to infinity of (sin(n))/(n^{3/2)} is converges (absolutely)