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.)
['https://math.stackexchange.com/questions/2704534', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/544878/']
Your approach is correct: $a\_n$ is positive and by applying the root test the convergence of the series $\sum\_n (1/a\_n)^{n}$ follows as soon as you show that $1/a\_n\to L<1$. Here we have that $1/a\_n\to 0$ so $\sum\_n (1/a\_n)^{n}$ converges.

In fact, $\sin(t)\geq 2t/\pi$ for $t\in [0,\pi/2]$ ($\sin(x)$ is concave in that interval). Hence for $n>1$,
$$a\_n = \int\_1^n \sin{\left(\frac{1}{\sqrt{x}}\right)} \,dx\geq \frac{2}{\pi}\int\_1^n \frac{1}{\sqrt{x}}\, dx =\frac{4(\sqrt{n}-1)}{\pi}\implies 0<\frac{1}{a\_n}\leq \frac{\pi}{4(\sqrt{n}-1)}$$
which implies that $1/a\_n\to 0$ as $n$ goes to infinity.

P.S. By the way $\sin(t)\leq t$ for $t\geq 0$ implies that
$$a\_n = \int\_1^n \sin{\left(\frac{1}{\sqrt{x}}\right)} \,dx\leq \int\_1^n \frac{1}{\sqrt{x}}\, dx ={2(\sqrt{n}-1)}\implies \frac{1}{a\_n}\geq \frac{1}{2(\sqrt{n}-1)}$$
which implies that $\sum\_n (1/a\_n)=+\infty$.
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.