convergence or divergence of series?

Question

Sigma (1 + sin n^2)/ sqrt n , n = 1, to 00

Does this series converge or not?

I am stuck on this one and I am not convinced of the previuos answer so I ask again hoping there are more mathematicians online. Thanks in advance.
Does the comparison test work here? If so, what is the inequality.
I know -1 < = sin n^2 < = 1
0 < = 1 + sin n^2 < = 2
making our an < = 2/ sqrt n, but that is useless. I am trying to get a uselful comparison, can someone help?

Thank you Steiner. I did not know about this form.
I will do as you suggested and will get back to you if I get an answer.

Answer 1

For various reasons I am sure it diverges, but the question is, why? Well, suppose you could show sin n^2 is bounded away from -1 (and this is not true, but stay with me) then you would be done. In this case 1+sinn^2 > epsilon for some epsilon > 0 and then the series is
> sum epsion/sqrt(n)

Suppose you could show something weaker, for some epsilon that "most" of the time 1 + sin n^2 > epsilon. But that, I mean

let Sn = {k | 1 + sin k^2 > epsilon and k <= n}, then
Sn/n > .9 for all large n.

Let S = {k | 1 + sin k^2 > epsilon}.

If this is true, then the sum you are interested over all n is bigger than the sum over S. And the sum over S is greater than

Sigma _S epsilon/sqrt(n). Now this diverges because the of the construction of S and the fact this series is monotonic.

(This is a bit tricky, but you can ignore the start of any series and focus on the tail. If this series converges then the sum over the compliment of S must diverge. But, with some work you can show the tail of the sum over S is bigger than the tail over the compliment of S -- mainly because there are so many more terms and the series is monotonic.)

So we would be done if we could show that for a small number epsilon, Sn/n > .9 for all large n.

Let C_n be the positive numbers less than or equal to n that are not in S_n.

Now there is a theorem that says polynomials will be uniformly distributed on a circle. The proof is kind of messy. There may well be a simple solution for n^2 and 10 years ago when I was in g-school I think I knew it -- but not enough time to think about it now. I will try and think of something.

Answer 2

Ok this is a tricky (sort of) series so I'm not 100% sure this works.

0 < = 1 + sin n^2 < = 2, you can compare the sum to:

Sigma (1 + sin n^2)/ n

since:

Sigma (1 + sin n^2)/ n < Sigma (1 + sin n^2)/ sqrt n

use the improper integral test.

integral from 1 to infinity [(1 + sin x^2)/ x] dx

= integral from 1 to infinity [1/x + sin (x^2)/x] dx

It is well known that the former part diverges to positive infinity, while the second part can be rewritten as a sine integral:

(1/2) integral from 0 to x^2 [sin t / t] dt

and it is well-known that

integral from 0 to infinity [sin t / t] dt = pi/2
and
integral from 0 to 1 [sin t / t] dt is a finite number.

Therefore integral from 1 to infinity [sin(x^2)/x] dx converges to a finite number.
But integral from 1 to infinity [1/x] dx diverges, which makes the whole integral diverge. Therefore the series diverges.

thats just my 2 cents


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Okay you're right Shane. Another possible approach would be to use Dirichlet's test but I will leave it to someone else.

Answer 3

I have no proof but I highly suspect that this series would diverge. The sin(n^2) behaves like a random variable centered at 0 and 1/ sqrt(n) diverges really quickly.

Also the improper integral test used in the previous answer doesn't fit the requirements of the integral test "non-negative monotone decreasing" and therefore doesn't constitute a working proof. Sorry.

Answer 4

Converges; limit is 2.

Answer 5

I thought about this problem, but, unfortunately, couldn't come to a conclusion. I suggest you should post this question to a more advanced Math forum. I strongly suggest Google sci.math. There are a alot of good mathematicians there. You'l probably get a good answer.

I would post this to sci. math, but I'm at work and my computer here can't t access that group.

If you get a good answer, please, let me know.

Steiner