Need help with a (apparently) difficult series

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Homework Statement
I have a series, but i cannot find a method to study the convergence. Can anyone, please, help me with this series? I can't understand what kind of method can be used to study its convergence.
Relevant Equations
I thought about using ##0\leq\left|\sin(n)\right|\leq1##
This is the series: $$\sum_{n=1}^{+\infty}\sin(n)\sin\left(\frac{1}{n}\right)\left(\cos\left(\frac{1}{\sqrt{n}}\right)-1\right)$$
 
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Good start. I'm sure that you will need to use the bound on ##\sin(n)## that you mention, but you need more than that. I think it must have something to do with the small angle approximations. For large ##n##, ##\sin(1/n) \approx 1/n## and ##\cos(1/\sqrt{n}) \approx 1- \frac{1}{2n}##. These approximations can be derived from the Taylor series expansions. But I have concerns about using approximations in an infinite series since I do not know how fast the terms approach their approximate value.
 
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...
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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