Convergence of $\sum_{n=2}^{\infty} a_n^{-n}$

Question

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.)

Answer 1

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 $\sum_n (1/a_n)=+\infty$.

Answer 2

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.

Answer 3

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.