Need help with an exercise on a sequence with integrals.

Question

Let $(I_{n})_{n \in \mathbb{N}}$ defined by $\forall n \in \mathbb{N}, I_{n}= \displaystyle \int_{0}^{\pi/4}\tan^n(x)dx.$

I proved that $(I_{n})_{n \in \mathbb{N}}$ decreases.

I'd like to prove that $\forall n \geq 2$, $\frac{1}{2(n+1)} \leq I_{n} \leq \frac{1}{2(n-1)}$.

Can anyone help me get started please? Thanks a lot.

Maybe if you integrate by parts you can express $I_{n+2}$ in terms of $I_{n+1}$ or $I_{n}$. A similar thing works for integral of $\sin^n x$. — D. Ungaretti, Sep 01 '19 at 22:01
I already proved that $\forall n \in \mathbb{N}, I_{n}+I_{n+2}= \frac{1}{n+1}$. — Pablito, Sep 01 '19 at 22:08
I started the same way you did, @Pablito. Then I proceeded to try induction: It worked fine for the upper inequality but not for the lower. :( — Ted Shifrin, Sep 01 '19 at 22:08
I went looking for other solutions and I found this: How to show the inequality for $\int_{0}^{\pi/4}\tan^n(x) dx$ — Trevor Gunn, Sep 01 '19 at 23:09

Trevor Gunn · Accepted Answer · 2019-09-01T22:24:14.913

1

I think you want to do something like $\tan(x) \ge x$ so

$$ I_n \ge \int_0^{\pi/4} x^n \,dx = \frac{1}{n + 1} \left(\frac{\pi}{4}\right)^{n+1}. $$

And $(\pi/4)^{2 + 1} < 1/2$. Then you do something similar for the upper bound.

edited Sep 01 '19 at 22:24

answered Sep 01 '19 at 22:06

Trevor Gunn

Oh yeah it works but I can't find any relation for the upper bound though :/ – Pablito Sep 01 '19 at 22:51
@Pablito I thought of using $(4/\pi)x \ge \tan(x)$ which gives $1/(n+1) \ge I_n$ but this isn't quite sharp enough. You can use the relation $I_n + I_{n + 2} = 1/(n+1)$ to get $I_{n+2} \le 1/(n + 1) - 1/2(n+1) = 1/2(n+1)$. This is what you want but it's only for $n + 2 \ge 4$ so you need to calculate $n = 2$ and $n = 3$ separately. This is what Ted mentioned in the comments. – Trevor Gunn Sep 01 '19 at 22:59

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