Does sin(4/3n) Converge or Diverge?

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given the series, determine whether the series converges or diverges:
(E is my sigma)

E sin(4/3n)


Now would it just be the same if i compared 1/n to 3/4n to diverge

Then could I just say since 3/4n diverges then sin(3/4n) must also diverge?

I wouldn't really know how to do the algebra for the direct comparison test or limit comparison test with the sin being there. But i can do the algebra for comparing 1/n with 3/4n..help please
 
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Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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