Why does the integral test not apply to #Sigma (sinn/n)^2# from #[1,oo)#?

Question

Elijah Abram · Accepted Answer

Because #f(x) =(sinx/x)^2# is not decreasing for #x in [1,+oo)#

The integral test for convergence of a series requires that the function #f(x)# such that #f(n) = a_n# has to be positive, decreasing and infinitesimal as #x->oo#

If we choose naturally:

#f(x) =(sinx/x)^2#

we can see that #f(x) is positive and infinitesimal, but not decreasing, so the conditions for the integral test are not met.

Christopher Agresta · Answer

undefined The integral test applies to series of the form $ \sum_{n=1}^{\infty} a_n $ where $ a_n $ is a positive, continuous, and decreasing function. However, for the series $ \sum_{n=1}^{\infty} \left(\frac{\sin n}{n}\right)^2 $, the function $ \left(\frac{\sin n}{n}\right)^2 $ does not meet the conditions required for the integral test. Specifically, it fails to meet the condition of being continuous and decreasing over the entire interval $[1, \infty)$. Since the function $ \frac{\sin n}{n} $ oscillates between positive and negative values for different integer values of $ n $, it does not satisfy the conditions for the integral test. Therefore, the integral test cannot be applied to this series.