Why does the series [(sin n)^2]/n^(1/2) diverge?

[Rudeboy37]

Homework Statement

Sum from 1 to infinity of [(sin n)^2]/n^(1/2)

Homework Equations

The Attempt at a Solution

I've tried every basic series test:

Test for Divergence: the limit approaches 0 so that doesn't tell us anything

Direct Comparison test: its smaller than n^(-1/2) which diverges so that doesn't tell us anything

Limit Comparison Test: I tried comparing it with both (sin n)^2 and n^(-1/2). The former's limit equals 0 which doesn't tell us anything since (sin n)^2 diverges by test for divergence. The latter's limit does not exist, thus making this test futile.

Ratio test/root test: just gets messy

The only other test I tried was the integral test. This function is a not a "nice" integrable function, but according to Wolfram-Alpha, the integral from 1 to infinity diverges (it mentioned the cauchy-principal value, so I'm assuming it's somehow using complex analysis) so I suppose this series would diverge. However, I am not a fan of proofs by Wolfram-Alpha, so I was wondering if anyone could help me figure out either a reason the integral diverges or a nicer way to show the series diverges (assuming I'm right and it does diverge).

 

[Char. Limit]

Well, first let me tell you that I proved (by Wolfram-Alpha) that the sum converges to a value of about 35. However, I'm as stumped as you are as to how we can prove that it converges.

 

[losiu99]

Could you please elaborate on how did you do that? Mathematica says the sum of first 10K elements is almost 100.
As for proving divergence: I'd try exploiting the fact that the sequence
\left{\frac{n}{2\pi}\right}
is equidistributed modulo 1. It means that for each subinterval, number of elements that fall into it is roughly equal to it's length. This allows us to say something like "for about 1/3 integers, \sin^2n>1/2, and so the series diverges, because
\sum_{n=1}^\infty\frac{1}{2\sqrt{3n}}
diverges.
Edit: it seems \sqrt{n} is a very good approximation of this sum.

 

[Char. Limit]

http://www.wolframalpha.com/input/?fp=1&i=sum_(n=1)^infinity+sin^2(n)/sqrt(n)&s=46&incTime=true

I used that.

 

[losiu99]

On the other hand,
http://www.wolframalpha.com/input/?i=sum_%28n%3D1%29^2000+sin^2%28n%29%2Fsqrt%28n%29

 

[Char. Limit]

Ah, I guess Wolfram-alpha lied to me.

 

[Rudeboy37]

And come to think of it, my integral test doesn't work since this is not decreasing. Losiu99, where does the sqrt3 come from?

 

[losiu99]

On the second thought, it shouldn't be there. Equidistribution doesn't guarantee that much regularity The best we can say knowing that 1/3 of sin^2n is not less than 1/2 is that the whole sum is not less than
1/2\sum_{n=1}^{N/3}\frac{1}{\sqrt{n}}

Edit: the actual numbers are pulled out of hat, they were meant only as an example.

 

[lurflurf]

It diverges because 1/n^(1/2) diverges, and all the [(sin n)^2] does is drop a few terms. Whenever a term is small the previous and next terms are not. Consider
[sin n]^2/n^(1/2) +[(sin (n-1)]^2/(n-1)^(1/2)
which converges or diverges with the original series and the terms of the new series cannot be small when compared to 1/n^(1/2).