Alternating Series Convergence: Is the Series Sum of sin(1/n^2) Convergent?

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Homework Statement



Sorry I don't know how to use symbols on this site so bear with me:

the question is does the following series converge: sum of sin(1/n^2) where n goes from 1 to infinity

Homework Equations



Limit comparison test, maybe others

The Attempt at a Solution



Okay I think I got the solution but I'm not sure if the logic is correct. I divided sin(1/n^2) by the p-series (1/n).

This comes out to n/sin(1/n^2)...the top will go to infinity and the bottom will go to 0 as n goes to infinity, so the series divergies. Because 1/n also diverges (harmonic series) sin(1/n^2) also diverges. Is this right?
 
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Don't know why I said 'alternating' series in the thread title...my brain is becoming mush!
 
I think limit test is probably the way to go, however I think you're doing a few things wrong with it.

With a_n being the given function, consider b_n = \frac{1}{n^2}

Then,
\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} n^2\sin{\frac{1}{n^2}}

Now, you should be able to show that that limit is finite, so the convergence of the sum of b_n would imply the convergence of the sum of a_n. Does that help?
 
Yep that fixes it, I also made an algebraic error thinking sin was on the denominator. Brain is going to mush. Thanks!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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