Does the following converge or diverge?
$$
\sum\_{n=2}^{\infty} a\_n^{-n},
$$
where$$
a\_n = \int\_1^n \sin{\left(\frac{1}{\sqrt{x}}\right)} \,\mathrm{d}x.
$$
My friends thought this sum would converge. I think we should do the square root test, checking the value of $t=\lim\limits\_{n \to \infty} \dfrac{1}{a\_n}$. And we knew that this goes to $0$. So, that would make the sum to converge…

Is there any problem in my ideas? Some said that this sum diverges, which I can never understand why…

It will be great if someone can explain me on this… Better with some proofs I can understand. (We are students learning calculus.)
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From $\sin x<x$, and for $n>4$,

$$\left(\int\_1^n\sin\frac1{\sqrt x}dx\right)^{-n}<\left(2\sqrt n\right)^{-n}<\frac1{n^2}$$

and the series converges.
As $n\to +\infty$ you have that asymptotically

$$\int\_1^n\sin\frac{1}{\sqrt{x}} \sim 2\sqrt{n}$$

Thence the related series goes like $\frac{1}{n^{n+1/2}}$ which converges.