How do you test the convergence of sigma with n = 1 to infinity of sin²x / (n sqroot n) ?

Answer 1

Well, first of all sin(x) is bounded between -1 and 1. Which means that sin(x)^2 is always between zero and one.

Therefore your series can be compared with 1/n sqrt(n).

Notice that 1/n sqrt(n) is always bigger than or equal to to sin(x)^2/n sqrt(n).

The sum 1/n sqrt(n) converges because of the p-series test.

Therefore your original series also converges.

Answer 2

The sin^2 x part doesn't matter since it's independent of n, thus you only need to consider whether the sum of 1/(n sqrt n) converges. This is the same as 1/n^{3/2}. As you may know the sum of 1/n^p converges when p>1. You can see this by considering the integral

\int_1^infty 1/n^p dn = 1/(p-1).

Answer 3

Note that sin² x is constant w.r.t. n, so [n=1, ∞]∑sin² x/(n√n) = sin² x [n=1, ∞]∑1/(n√n), which, like all series of the form [n=1, ∞]∑1/(n^s) with s>1, converges. (You may prove this with the integral test -- 1/n^s is monotone decreasing, so [n=1, ∞]∑1/n^s converges if [1, ∞]∫1/x^s dx does. But for s>1, 1-s<0, so [1, ∞]∫1/x^s dx = [1, ∞]∫x^(-s) dx = x^(1-s)/(1-s) | [1, ∞] = [h→∞]lim h^(1-s)/(1-s) - 1^(1-s)/(1-s) = 1/(s-1)).

Answer 4

Well since (sin(x))^2 oscillates between -1 and 1, and (n(sqrt(n))) is equivalent to n^(3/2) which grows rapidly as n-> oo, 1/oo=0.