Question

How would I determine if this series converges or diverges: sum from 1 to infinity of (sin(1/n^2))

Answer 1

Yeah

Answer 2

\[\sum_{n=1}^{\infty} \sin(\frac{ 1 }{ n^2 })\]

Answer 3

sin(1/n^2 )

Answer 4

use\[\sin x<x\]on \((0,\infty)\)

Answer 5

so sin(1/x^2) is less than 1/x^2 for all x > 0, 1/x^2 is a convergent p-series, thus sin(1/x^2) converges by Comparison Test?

Answer 6

yep :)

Answer 7

nice! thanks :)

Answer 8

welcome!